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4. -JC' + 2/^ + 2</a; + 2fy + c = which are parallel to the line

a; + 22/-6 = 0.

5. Prove that the straight line y = x + cJ2 touches the circle

x'^-\-y'^ = c-\ and find its point of contact.

6. Find the condition that the straight line ex ~hy + 1x^ = may

touch the circle x'^ + y^ = ax + by and find the point of contact.

7. Find whether the straight line x + y = 2 + J 2, touches the circle

x- + y^-2x-2y + l = Q.

8. Find the condition that the straight line ^x + ^y = k may

touch the circle x'^-\-y'^ â€” lQx,

9. Find the value oip so that the straight line

cccos a+?/sina-p =

may touch the circle

a;2 + 7/2 - 2aa; cos a - 2hy sin a - a^ sin^ a â€” 0.

10. Find the condition that the straight line Ax-\-By + G â€” Q may

touch the circle

{x-af+{y-hf=c^.

11. Find the equation to the tangent to the circle x - \-y'^ â€” a?

which

(i) is parallel to the straight line y = mx + c,

(ii) is perpendicular to the straight line y = mx + c,

(iii) passes through the point (6, 0),

and (iv) makes with the axes a triangle whose area is a-.

12. Find the length of the chord joining the points in which the

straight line

X y ,

a

meets the circle x'^ + y^ = r".

13. Find the equation to the circles which pass through the origin

and cut off equal chords a from the straight lines y â€” x and yâ€”-x.

[EXS. XVIII.] EXAMPLES. 135

14. Find the equation to the straight lines joining the origin to

the points in which the straight Hne y = rnx + c cuts the circle

x^ + y'^ = 2ax + 2by .

Hence find the condition that these points may subtend a right

angle at the origin.

Find also the condition that the straight line may touch the

circle.

Find the equation to the circle which

15. has its centre at the point (3, 4) and touches the straight line

5x + 12y = l.

16. touches the axes of coordinates and also the line

a

the centre being in the positive quadrant.

17. has its centre at the point (1, - 3) and touches the straight

Hne 2a: -i/- 4 = 0.

18. Find the general equation of a circle referred to two perpen-

dicular tangents as axes.

19. Find the equation to a circle of radius /â€¢ which touches the

axis of 2/ at a point distant h from the origin, the centre of the circle

being in the positive quadrant.

Prove also that the equation to the other tangent which passes

through the origin is

{r^-h'^)x + 2rhy = 0.

20. Find the equation to the circle whose centre is at the point

(a, p) and which passes through the origin, and prove that the

equation of the tangent at the origin is

21. Two circles are drawn through the points {a, 5a) and (4a, a)

to touch the axis of y. Prove that they intersect at an angle tan^^ y* .

22. A circle passes through the points ( - 1, 1), (0, 6), and (5, 5).

Find the points on this circle the tangents at which are parallel to the

straight line joining the origin to its centre.

160. To shew that from any i^oint there caii he drawn

two tangents^ real or imaginary^ to a circle.

Let the equation to the circle be '3t? â– \- y"^ â€” Â«', and let the

given point be (ajj, y^. [Fig. Art. 161.]

The equation to any tangent is, by Art. 155,

y â€” mx + a J\ + rn?.

136 COORDINATE GEOMETllY.

If this pass through the given point (xj^, y^ we have

v/i = mx'i + aj\ + m^ (1).

This is the equation which gives the values of m corre-

sponding to the tangents which pass through (x-^^^ y^).

Now (1) gives

?/i â€” jiid\ â€” a Jl + m^,

i.e. y^ â€” 2rax-^y-^ + m^x^ = Â«- + ahn^^

i. e. m^ {xj^ â€” a^) â€” 2mx^y^ + y-^^ â€” a^ = (2).

The equation (2) is a quadratic equation and gives

therefore two values of ?m (real, coincident, or imaginary)

corresponding to any given values of x^^ and y^. For each

of these values of 7?i we have a corresponding tangent.

The roots of (2) are, by Art. 1, real, coincident or

imaginary according as

(2xi?/i)^ â€” 4 (.r/ â€” a"^) (2/1^ â€” a^) is positive, zero, or negative,

i. e. according as

a^ (â€” a"^ + Xj^ + y^) is positive, zero, or negative,

i. e. according as x-^ + y^ =â– - a?.

If x^ + y^> a^, the distance of the point (x^^, y^) from

the centre is greater than the radius and hence it lies outside

the circle.

If x{- + yi = a?, the point {x^ , y^ lies on the circle and

the two coincident tangents become the tangent at (x-^^ , y^.

If x-^ + y^ <Â«^, the point {x-^^ y^ lies within the circle,

and no tangents can then be geometrically drawn to the

circle. It is however better to say that the tangents are

imaginary.

161. Chord of Contact. Def. If from any point

T without a circle two tangents TP and TQ be drawn to

the circle, the straight line PQ joining the points of

contact is called the chord of contact of tangents from T.

To find the equation of the chord of contact of tangents

drawn to the circle x^ + y"^ â€” a^ from the external point

(^1, 2/i)-

POLE AND POLAR.

137

â€¢(3),

.(4).

Let T be the point (x^, y^), and F and ^ the points

(oj'j 2/') and (a?", y") respectively.

The tangent at F is

XX +1/1/' â€” a-^ (1),

and that at Q is

xx" + yy" â€”d^ (2).

Since these tangents pass through

T^ its coordinates {x-^ , y^ must satisfy

both (1) and (2).

Hence x-^ + y^y â€” a?

and x^od' + y-^y" = cir

The equation to FQ is then

xxi + yyi=a2 (5).

For, since (3) is true, it follows that the point {x ^ y),

i.e. F, lies on (5).

Also, since (4) is true, it follows that the point (x\ y")^

i.e. Q, lies on (5).

Hence both F and Q lie on the straight line (5), i.e.

(5) is the equation to the required chord of contact.

If the point {x^, y^) lie within the circle the argument

of the preceding article will shew that the line joining the

(imaginary) points of contact of the two (imaginary)

tangents drawn from (x^ , y^) is xx^ + yy^ â€” a^.

We thus see, since this line is always real, that we may

have a real straight line joining the imaginary points of

contact of two imaginary tangents.

162. Pole and Polar. Def. If through a point

F (within or without a circle) there be drawn any straight

line to meet the circle in Q and F, the locus of the point of

intersection of the tangents at Q and F is called the polar

of F ; also F is called the pole of the polar.

In the next article the locus will be proved to be a

straight line.

138

COORDINATE GEOMETRY.

163. To Jincl the equation to the 'polar of the point

(^ij 2/1) '^^ih respect to the circle x^ â– \-y^ â€” d^.

Let QR be any chord drawn through P and let the

tangents at Q and R meet in the point :Z^ whose coordinates

are (A, k).

Hence QR is the chord of contact of tangents drawn

from the point (h, k) and therefore, by Art. 161, its

equation is xh â– \- yk â€” o?.

Since this line passes through the point (x-^, y^ we

have

x-Ji + y^ â€” a" (1).

Since the relation (1) is true it follows that the

variable point (A, k) always lies on the straight line whose

equation is

xxi + yyi = a2 (2).

Hence (2) is the polar of the point (x^, y^).

In a similar manner it may be proved that the polar of

(x'l, i/i) with respect to the circle

o(^ + y^ + 2gx + 2fy + c =

is xx^ -^yvi + g (x + xi) +/ (y + yi) + c=^ 0.

164. The equation (2) of the preceding article is the

same as equation (5) of Art. 161. If, therefore, the point

{xi, 2/i) be without the circle, as in the right-hand figure,

the polar is the same as the chord of contact of the real

tangents drawn through (x^^, y^).

If the point (x^, y^) be on the circle, the polar coincides

with the tangent at it. (Art. 150.)

GEOMETRICAL CONSTRUCTION FOR THE POLAR. 139

If the point (x^, y-f) be within the circle, then, as in

Art. 161, the equation (2) is the line joining the (imaginary)

points of contact of the two (imaginary) tangents that can

be drawn from (x-^ , y^.

We see therefore that the polar might have been

defined as follows :

The polar of a given point is the straight line which

passes through the (real or imaginary) points of contact of

tangents drawn from the given point ; also the pole of any

straight line is the point of intersection of tangents at the

points (real or imaginary) in which this straight line meets

the circle.

165. Geometrical construction for the imlar of a point.

The equation to OP^ which is the line joining (0, 0) to

^.e.

â– (!)â€¢

Also the polar of P is

xx^ + yVx â€” cC"

(2).

By Art. 69, the lines (1) and (2) are perpendicular to

one another. Hence OP is perpendicular to the polar

of P.

Also the length OP - ^x^+y},

140 COORDINATE GEOMETRY,

and the perpendicular, OiV, from upon (2)

Hence the product ON . OP â€” a^.

The polar of any point P is therefore constructed thus :

Join OP and on it (produced if necessary) take a point N

such that the rectangle ON . OP is equal to the square of

the radius of the circle.

Through N draw the straight line LL' perpendicular to

OP ; this is the polar required.

[It will be noted that the middle point N of any chord LL' lies on

the line joining the centre to the pole of the chord.]

166. To Jind the pole of a given line with respect to

any circle.

Let the equation to the given line be

^a; + % + (7-0 (1).

(1) Let the equation to the circle be

and let the required pole be {x-^, y-^.

Then (1) must be the equation to the polar of (oji, y^,

i.e. it is the same as the equation

xx-^ + yVi â€” cv^-O (2).

Comparing equations (1) and (2), Ave have

J B C '

so that Xt â€” â€” yz a^ and y, = â€” '^ a?.

C ^ C

The required pole is therefore the point

A , B ^

(2) Let the equation to the circle be

a;- + 2/2 + l.yx + 2fy + g=Q.

POLE AND POLAR. 141

If (a?!, 2/1) ^^ *^Â® required pole, then (1) must be

equivalent to the equation

xx^ +yyi + g{os + x,) +f{y + yi) + c=0, (Art. 1 63),

i.e. x(x^+g)+y{y^+f) + gxj^ +/2/i + c = (3).

Comparing (1) with (3), we therefore have

- sc^ + 9 _ Vi+f^ 9^1 +/2/1 + g

A " B C

By solving these equations we have the values of x^

and 2/1.

Ex. Find the pole of the straight line

9x+2j-28 = (1)

2vith respect to the circle

2x^+2y'^-Sx + 5ij-7 = (2).

If (.Tj, yj) be the required point the line (1) must coincide with the

polar of (x^, 2/i)' whose equation is

2xx^ + 2yy-^-i{x + x^)+^{y+y{j-l = 0,

i.e. x{4x^-d) + y{4:y^ + 5)-3xi + 5y^-U = (3).

Since (1) and (3) are the same, we have

4.'Ci - 3 _ 4|/i + 5 _ -Sxi + 5y^ - 14

9 r~ ~ ^28 â–

Hence Xi = 9yi + 12,

and 3.Ti-117i/i = 126.

Solving these equations we have Xj^ = S and t/^ = - 1, so that the

required point is (3, - 1).

167. If the 2')olar of a point P pass through a point T,

then the polar of T passes through P.

Let P and T be the points {x-^, y^ and (.Tg, 2/2) ^^-

spectively. (Fig. Art. 163.)

The polar of (x^, y^ with respect to the circle

a;^ + 2/2 _ ^2 jg

xx-^ + 2/2/1 = Cb^'

This straight line passes through the point T if

jcoa?! + 2/22/1 = ct^ â€¢â€¢â€¢ (!)â€¢

142 COORDINATE GEOMETRY.

Since the relation (1) is true it follows that the point

(x^, 2/1), i.e. P, lies on the straight line xx^-\-yy^ = (]?^ which

is the polar of {x^^ y^)^ i.e. T, with respect to the circle.

Hence the proposition.

Cor. The intersection, T, of the polars of two points,

P and Q, is the pole of the line PQ.

168. To find the length of the tangent that can he

drawn from the point (x^ , y-^ to the circles

(1) a?-^y'^ = a\

and (2) x? + y^ + 2gx + 2/^/ + c = 0.

If T be an external point (Fig. Art. 163), TQ a tangent

and the centre of the circle, then TQO is a right angle

and hence

(1) If the equation to the circle be x" ^-y"- â€” a^, is the

origin, OT'^ = x^- + y^.^ and OQ^ = a^.

Hence TQ" = x^^ + y^- - a\

(2) Let the equation to the circle be

x^ + y'^+ 2gx + 2fy + c = 0,

i.e. (x + gf-v{y+fY = f+f'-c.

In this case is the point (â€” g, â€”f) and

OQ^ â€” (radius)^ = g'^ +f^ â€” c.

Hence OT^ = [x, - (- g)Y + [y, - (-/)]' (Art. 20).

Therefore TQ^ ^ {x, + gf + (y, +ff - (f +p - c)

= <^ + Vi+'^g^i + 2/J/i + c.

In each case we see that (the equation to the circle

being written so that the coefficients of x^ and y" are each

unity) the square of the length of the tangent drawn to the

circle from the point {x^, y^) is obtained by substituting x^

and 2/1 for the current coordinates in the left-hand member

of the equation to the circle.

*169. To find the equation to the pair of tangents that

can he drawn from the point (x-^^ y^) to the circle x- + y^ â€” d^.

PAIR OF TANGENTS FROM ANY POINT. 143

Let (A, k) be any point on either of the tangents from

(^1, Vi)'

Since any straight line touches a circle if the perpen-

dicular on it from the centre is equal to the radius, the

perpendicular from the origin upon the line joining (x^, y^)

to (A, k) must be equal to a.

The equation to the straight line joining these two

points is

^ â€” Vi /

i.e. yQ^~~ ^'i) ~ ^ (^ ~ 2/1) + ^^1 ~~ ^^Ui â€” ^â–

__ kx^ â€” hy^

Hence , ^ = a,

J(h-x,f+(k-y,y

so that (kx^ â€” hy^y = a^ [{h â€” oo^y + {k â€” 2/1)^]-

Therefore the point (h, k) always lies on the locus

{x^y - xy^f = a" [{x - x^y + {y- y,y] (1).

This therefore is the required equation.

The equation (1) may be written in the form

= 2xyx^yi â€” 2a^xXj^ â€” 2a^yyi,

i.e. (x^ + y'^- CL^) {^i + y\ â€” Â«^) = x'^X{' + y^y^ + a^ + 2xyx{y^

â€” 2a^xXi â€” 2c^yy-^ â€” {xx^ + yy^ â€” a^y (2).

#170. In a later chapter we shall obtain the equation to the pair

of tangents to any curve of the second degree in a form analogous

to that of equation (2) of the previous article.

Similarly the equation to the pair of tangents that can be

drawn from {x-^, y-^ to the circle

If the equation to the circle be given in the form

a;2 + 2/2 + 2,9ra; + 2/?/ + c =

the equation to the tangents is, similarly,

{x' + ^2 + ^g^ + 2/y + c) {x^ + y^ + 2^a;i + 2jy^ + c)

= \xx^^yy^^g{x\x^^f{y^ry^) + cf (2).

144 COORDINATE GEOMETRY.

EXAMPLES. XIX.

Find the polar of the point

1. (1, 2) with respect to the circle a;- + y^ = 7.

2. (4, - 1) with respect to the circle 2x" + 2y^=ll,

3. (-2,3) with respect to the circle

x'^ + y^-4x-&y + 5 = 0.

4. (5, - I) with respect to the circle

Sx^ + Sy^-1x + 8y-9 = 0.

5. (a, - 6) with respect to the circle

x^ + y^ + 2ax - 2by + a^ - b- = 0.

Find the pole of the straight line

6. x + 2y = l with respect to the circle x^ + y^=5.

7. 2x-y = G with respect to the circle 5x^ + oy^ =9.

8. 2x + y + 12 = with respect to the circle

x^ + y^-4x + By-l=0.

9. 48;r - 54?/ + 53 = with respect to the circle

Sx^ + 3y^ + 5x-7y + 2 = 0.

10. ax + by + 3a^ + 36^ = with respect to the circle

x^ + y^ + 2ax + 2by = a^ + b^.

11. Tangents are drawn to the circle x^ + y^ = 12 at the points

where it is met by the circle x^ + y^ - 5x + Sy ~ 2 = ; find the point of

intersection of these tangents.

12. Find the equation to that chord of the circle x'^ + y^ = 81 which

is bisected at the point ( - 2, 3), and its pole with respect to the circle.

13. Prove that the polars of the point (1, - 2) with respect to the

circles whose equations are

x^ + y^ + Qy + 5 = and x'^ + y^ + 2x + 8y + 5 =

coincide ; prove also that there is another point the polars of which

with respect to these circles are the same and find its coordinates.

14. Find the condition that the chord of contact of tangents from

the point (x', y') to the circle x^+y^=a^ should subtend a right angle

at the centre.

15. Prove that the distances of two points, P and Q, each from

the polar of the other with respect to a circle, are to one another

inversely as the distances of the points from the centre of the circle.

16. Prove that the polar of a given point with respect to any one

of the circles x^-\-y'^-2kx-\-c'^ = 0, where k is variable, always passes

through a fixed point, whatever be the value of A\

[EXS. XIX.] POLAR EQUATION TO THE CIRCLE.

145

17. Tangents are drawn from the point {h, k) to the circle

x^ + y^ = aP; prove that the area of the triangle formed by them

and the straight line joining their points of contact is

ajJi^ + Ji^-aY

h^ + k^

Find the lengths of the tangents drawn

18. to the circle '2x^ + 2y'^â€”S from the point ( - 2, 3).

19. to the circle 3x^ + 3y^ -7x-6y = 12 from the point (6, - 7).

20. to the circle x^ + f' + Ihx - 36^ = from the point

(a + &, a-h).

21. Given the three circles

3ic2 + 37/-36a; + 81 = 0,

and a:2^.,y2_i6^_12y + 84 = 0,

find (1) the point from which the tangents to them are equal in

length, and (2) this length.

22. The distances from the origin of the centres of three circles

ic2-hy"^-2Xic = c^ (where c is a constant and X a variable) are in

geometrical progression ; prove that the lengths of the tangents drawn

to them from any point on the circle a;"^ + ^- = c-are also in geometrical

progression.

23. Find the equation to the pair of tangents drawn

(1) from the point (11, 3) to the. circle xr'-\-y^ â€” ^^,

(2) from the point (4, 5) to the circle

2x - + 2i/3 - 8a; + 12?/ + 21 r= 0.

171. To jincl the general equation of a circle referred

to polar coordinates.

Let be the origin, or pole, OX the initial line, G the

centre and a the radius of the

circle.

Let the polar coordinates of C

be R and a, so that 00 â€” M and

L XOC = a.

Let a radius vector through

at an angle 6 with the initial line

cut the circle in P and Q. Let

OP, or OQ, be r.

L.

10

146 COOKDINATE GEOMETRY.

Then {Trig. Art. 164) we have

CP- ^ OC'^ -r OP' -20C. OP cos C OF,

i.e. a''^P^ + r'-2Prcos(e~a),

i.e. r'- -2Rr cos {0 - a) + R"" - a" ^0 (1).

This is the required polar equation.

172. Particular cases of the general equation inpolar coordinates.

(1) Let the initial line be taken to go through the centre G. Then

a = 0, and the equation becomes

r2 - 2Rr cos ^ + E^ - a^ = 0.

(2) Let the pole be taken on the circle, so that

B=OG = a.

The general equation then becomes

^^M r^-2arcos{d-a) = 0,

i.e. r=2acos{d - a).

(3) Let the pole be on the circle and also let the initial line pass

through the centre of the circle. In this case

a=0, and R â€” a.

The general equation reduces then to the

simple form r = 2acos^.

This is at once evident from the figure. 0|

For, if OCA be a diameter, we have

0P=0^ cos ^,

i.e. r=2aQO8 0.

173. The equation (1) of Art. 171 is a quadratic

equation which, for any given value of 6, gives two

values of r. These two values in the figure are OP and

OQ.

If these two values be called r-^ and rg, we have, from

equation (1),

7\r^ = product of the roots â€” E^ â€” a"^,

i.e. OP.OQ^P'-a\

The value of the rectangle OP . OQ is therefore the

same for all values of 0. It follows that if we drew any

other line through to cut the circle in P^ and Q^ we

should have OP . OQ = OP^ .OQ^.

This is Euc. iii. 36, Cor.

POLAR EQUATION TO THE TANGENT. 147

174. Find the equation to the chord joining the points on the circle

r â€” 2a cos 6 whose vectorial angles are d^ and 6^, and deduce the equation

to the tangent at the point 6^.

The equation to any straight line in polar coordinates is (Art. 88)

^ = rcos((9-a) (1).

If this pass through the points (2a; cos ^j^, 6^ and (2asin^2Â» ^2)' ^^^

have

2a cos d-^ cos (^j-a)=^ = 2acos ^.3 cos {9.^~a) (2).

Hence cos (2^^ - a) + cos a = cos (2^., - a) + cos a,

i.e. 2^1 -a= -(2^0 -a),

since d^ and ^2 ^^^ iiotÂ» i'^ general, equal.

Hence a â€” d-^ + d.^^

and then, from (2), j? = 2a cos 6-^ cos d.-^.

On substitution in (1), the equation to the required chord is

r cos (^ - ^1 - ^2) = 2a cos ^1 cos ^2 (3).

The equation to the tangent at the point d^ is found, as in

Art. 150, by putting 6^ = 6^ in equation (3).

We thus obtain as the equation to the tangent

rcos (^-2^^) â€” 2acos2^j^. â€¢

As in the foregoing article it could be shewn that the equation to

the chord joining the points d^ and 6^ on the circle r= 2a cos {d - 7) is

r cos {d -9-^- 6., + 7] â€” 2a cos {9-^ - 7) cos (9.^ - 7)

and hence that the equation to the tangent at the point ^^ is

r cos (^ - 2^j + 7) = 2a cos2 (^j - 7).

EXAMPLES. XX.

1. Find the coordinates of the centre of the circle

r=A cos 9 + B sin 9.

2. Find the polar equation of a circle, the initial line being a

tangent. What does it become if the origin be on the circumference?

3. Draw the loci

(1) r=a; (2) r = a sin ^; (3) r = ttcos^; (4) r = asec^;

(5) r = acos(^-a); (6) r = asec(^-a).

4. Prove that the equations r = acos(^-a) and râ€”b sin (9 -a)

represent two circles which cut at right angles.

5. Prove that the equation r^ cos ^- a;* cos 2^ -2a- cos ^ =

represents a straight line and a circle.

10â€”2

148 COORDINATE GEOMETRY. [Exs. XX.]

6. Find the polar equation to the circle described on the straight

line joining the points {a, a) and (&, /3) as diameter.

7. Prove that the equation to the circle described on the straight

line joining the points (1, 60Â°) and (2, 30Â°) as diameter is

r^-r [cos {d - 60Â°) + 2 cos {0 - 30Â°)] + ^3 = 0.

8. Find the condition that the straight line

- = acos d + h sin^

r

may touch the circle r = 2c cos d.

175. To find the general equation to a circle referred to

oblique axes which meet at an angle to.

Let C be the centre and a the radius of the circle. Let

the coordinates of C be (h, k) so

that if CM, drawn parallel to the

axis of 2/, meets OX in M, then

OM=h and MC=.k.

Let P be any point on the

circle whose coordinates are x and

y. Draw PN, the ordinate of P,

and CL parallel to OX to meet

PNinL.

Then CL = MN = OiV - OM =- x - h,

and LP^NP-NL^NP -MG^y- k.

Also z CLP - z ONP= 180Â° - z PNX = 180Â° - tu.

Hence, since CL^- + LP^ - WL . LP cos CLP = a\

we have (x - h)2 + (y - k)2 + 2 (x - h) (y - k) cosco = 3?,

i.e. x^ + y^ + 2xy cos w - 2x [h + k cos w) ~2y ik + h cos w)

+ JiT + k^ + 2M COS w â€” or.

The required equation is therefore found.

176. As in Art. 142 it may be shewn that the

equation

aj- + ^xy cos oi-\- y" + '2gx + 2fy + c -

represents a circle and its radius and centre found.

OBLIQUE COORDINATES. J 49

Ex. If the axes he inclined at 60Â°, prove that the equation

x^ + xy+i/-^x-5y -2 â€” (1)

represents a circle and find its centre and radius.

If w be equal to 60Â°, so that cosw=|, the equation of Art. 175

D6COII1GS

x'^ + xy->ry - x{2h + Tc)-y{2k + h) + h'^ + l'^ + hh = a:^.

This equation agrees with (1) if

2/i + /c = 4 (2),

'ih + h^b (3),

and 7i2 + A;2 + 7i;e-a2= -2 (4).

Solving (2) and (3), we have h = l and k â€” 2. Equation (4) then

gives

- a2^7t2+F + ;iA; + 2z^9,

so that a = 3.

The equation (1) therefore represents a circle whose centre is the

point (1, 2) and whose radius is 3, the axes being inclined at 60Â°.

EXAMPLES. XXI.

Find the inclinations of the axes so that the following equations

may represent circles, and in each ease find the radius and centre ;

1. x'-xy-\-y'^~2gx-'ify = 0.

2. x'^ + fjZxy+if-^x-%y + 5 = Q.

3. The axes being inclined at an angle w, find the centre and

radius of the circle

a;2 + 2xy cos w + t/^ - Igx - 2fy = 0.

4. The axes being inclined at 45Â°, find the equation to the circle

whose centre is the point (2, 3) and whose radius is 4.

5. The axes being inclined at 60Â°, find the equation to the circle

whose centre is the point ( - 3, - 5) and whose radius is 6.

6. Prove that the equation to a circle whose radius is a and

which touches the axes of coordinates, which are inclined at an angle

to, is

x^ + 2xycoso} + y'^-2a (.T + ^)cot- +a^cot^-=0.

7. Prove that the straight line y â€” mx will touch the circle

x^ + 2xy cos (o + y^ + 2gx + 2fy + c =

if {g +fm)^ = c (1 + 2m cos o) + m^) .

8. The axes being inclined at an angle w, find the equation to the

circle whose diameter is the straight line joining the points

{x\ y') and {x", y").

150 COORDINATE GEOMETRY.

Coordinates of a point on a circle expressed in

terms of one single variable.

177. If, in the figure of Art. 139, we put the angle

MOP equal to a, the coordinates of the point P are easily

seen to be a cos a and a sin a.

These equations clearly satisfy equation (1) of that

article.

The position of the point P is therefore known when

the value of a is given, and it may be, for brevity, called

" the point a."

With the ordinary Cartesian coordinates we have to

give the values of two separate quantities x and y' (which

are however connected by the relation x' = Ja^ â€” y"^) to

express the position of a point P on the circle. The

above substitution therefore often simplifies solutions of

problems.

178. To find tlie equation to the straight line joining

two joints, a and y8, 07i the circle xr + y^ = a-.

Let the points be P and Q, and let OA^ be the perpen-

dicular from the origin on the straight line PQ ; then OJV

bisects the angle POQ, and hence

z X0]^= 1 ( z XOP + L XOQ) = HÂ« + ^)-

Also OK^ OP cos NOP = a cos Â°^^ .

The equation to PQ is therefore (Art. 53),

a; + 22/-6 = 0.

5. Prove that the straight line y = x + cJ2 touches the circle

x'^-\-y'^ = c-\ and find its point of contact.

6. Find the condition that the straight line ex ~hy + 1x^ = may

touch the circle x'^ + y^ = ax + by and find the point of contact.

7. Find whether the straight line x + y = 2 + J 2, touches the circle

x- + y^-2x-2y + l = Q.

8. Find the condition that the straight line ^x + ^y = k may

touch the circle x'^-\-y'^ â€” lQx,

9. Find the value oip so that the straight line

cccos a+?/sina-p =

may touch the circle

a;2 + 7/2 - 2aa; cos a - 2hy sin a - a^ sin^ a â€” 0.

10. Find the condition that the straight line Ax-\-By + G â€” Q may

touch the circle

{x-af+{y-hf=c^.

11. Find the equation to the tangent to the circle x - \-y'^ â€” a?

which

(i) is parallel to the straight line y = mx + c,

(ii) is perpendicular to the straight line y = mx + c,

(iii) passes through the point (6, 0),

and (iv) makes with the axes a triangle whose area is a-.

12. Find the length of the chord joining the points in which the

straight line

X y ,

a

meets the circle x'^ + y^ = r".

13. Find the equation to the circles which pass through the origin

and cut off equal chords a from the straight lines y â€” x and yâ€”-x.

[EXS. XVIII.] EXAMPLES. 135

14. Find the equation to the straight lines joining the origin to

the points in which the straight Hne y = rnx + c cuts the circle

x^ + y'^ = 2ax + 2by .

Hence find the condition that these points may subtend a right

angle at the origin.

Find also the condition that the straight line may touch the

circle.

Find the equation to the circle which

15. has its centre at the point (3, 4) and touches the straight line

5x + 12y = l.

16. touches the axes of coordinates and also the line

a

the centre being in the positive quadrant.

17. has its centre at the point (1, - 3) and touches the straight

Hne 2a: -i/- 4 = 0.

18. Find the general equation of a circle referred to two perpen-

dicular tangents as axes.

19. Find the equation to a circle of radius /â€¢ which touches the

axis of 2/ at a point distant h from the origin, the centre of the circle

being in the positive quadrant.

Prove also that the equation to the other tangent which passes

through the origin is

{r^-h'^)x + 2rhy = 0.

20. Find the equation to the circle whose centre is at the point

(a, p) and which passes through the origin, and prove that the

equation of the tangent at the origin is

21. Two circles are drawn through the points {a, 5a) and (4a, a)

to touch the axis of y. Prove that they intersect at an angle tan^^ y* .

22. A circle passes through the points ( - 1, 1), (0, 6), and (5, 5).

Find the points on this circle the tangents at which are parallel to the

straight line joining the origin to its centre.

160. To shew that from any i^oint there caii he drawn

two tangents^ real or imaginary^ to a circle.

Let the equation to the circle be '3t? â– \- y"^ â€” Â«', and let the

given point be (ajj, y^. [Fig. Art. 161.]

The equation to any tangent is, by Art. 155,

y â€” mx + a J\ + rn?.

136 COORDINATE GEOMETllY.

If this pass through the given point (xj^, y^ we have

v/i = mx'i + aj\ + m^ (1).

This is the equation which gives the values of m corre-

sponding to the tangents which pass through (x-^^^ y^).

Now (1) gives

?/i â€” jiid\ â€” a Jl + m^,

i.e. y^ â€” 2rax-^y-^ + m^x^ = Â«- + ahn^^

i. e. m^ {xj^ â€” a^) â€” 2mx^y^ + y-^^ â€” a^ = (2).

The equation (2) is a quadratic equation and gives

therefore two values of ?m (real, coincident, or imaginary)

corresponding to any given values of x^^ and y^. For each

of these values of 7?i we have a corresponding tangent.

The roots of (2) are, by Art. 1, real, coincident or

imaginary according as

(2xi?/i)^ â€” 4 (.r/ â€” a"^) (2/1^ â€” a^) is positive, zero, or negative,

i. e. according as

a^ (â€” a"^ + Xj^ + y^) is positive, zero, or negative,

i. e. according as x-^ + y^ =â– - a?.

If x^ + y^> a^, the distance of the point (x^^, y^) from

the centre is greater than the radius and hence it lies outside

the circle.

If x{- + yi = a?, the point {x^ , y^ lies on the circle and

the two coincident tangents become the tangent at (x-^^ , y^.

If x-^ + y^ <Â«^, the point {x-^^ y^ lies within the circle,

and no tangents can then be geometrically drawn to the

circle. It is however better to say that the tangents are

imaginary.

161. Chord of Contact. Def. If from any point

T without a circle two tangents TP and TQ be drawn to

the circle, the straight line PQ joining the points of

contact is called the chord of contact of tangents from T.

To find the equation of the chord of contact of tangents

drawn to the circle x^ + y"^ â€” a^ from the external point

(^1, 2/i)-

POLE AND POLAR.

137

â€¢(3),

.(4).

Let T be the point (x^, y^), and F and ^ the points

(oj'j 2/') and (a?", y") respectively.

The tangent at F is

XX +1/1/' â€” a-^ (1),

and that at Q is

xx" + yy" â€”d^ (2).

Since these tangents pass through

T^ its coordinates {x-^ , y^ must satisfy

both (1) and (2).

Hence x-^ + y^y â€” a?

and x^od' + y-^y" = cir

The equation to FQ is then

xxi + yyi=a2 (5).

For, since (3) is true, it follows that the point {x ^ y),

i.e. F, lies on (5).

Also, since (4) is true, it follows that the point (x\ y")^

i.e. Q, lies on (5).

Hence both F and Q lie on the straight line (5), i.e.

(5) is the equation to the required chord of contact.

If the point {x^, y^) lie within the circle the argument

of the preceding article will shew that the line joining the

(imaginary) points of contact of the two (imaginary)

tangents drawn from (x^ , y^) is xx^ + yy^ â€” a^.

We thus see, since this line is always real, that we may

have a real straight line joining the imaginary points of

contact of two imaginary tangents.

162. Pole and Polar. Def. If through a point

F (within or without a circle) there be drawn any straight

line to meet the circle in Q and F, the locus of the point of

intersection of the tangents at Q and F is called the polar

of F ; also F is called the pole of the polar.

In the next article the locus will be proved to be a

straight line.

138

COORDINATE GEOMETRY.

163. To Jincl the equation to the 'polar of the point

(^ij 2/1) '^^ih respect to the circle x^ â– \-y^ â€” d^.

Let QR be any chord drawn through P and let the

tangents at Q and R meet in the point :Z^ whose coordinates

are (A, k).

Hence QR is the chord of contact of tangents drawn

from the point (h, k) and therefore, by Art. 161, its

equation is xh â– \- yk â€” o?.

Since this line passes through the point (x-^, y^ we

have

x-Ji + y^ â€” a" (1).

Since the relation (1) is true it follows that the

variable point (A, k) always lies on the straight line whose

equation is

xxi + yyi = a2 (2).

Hence (2) is the polar of the point (x^, y^).

In a similar manner it may be proved that the polar of

(x'l, i/i) with respect to the circle

o(^ + y^ + 2gx + 2fy + c =

is xx^ -^yvi + g (x + xi) +/ (y + yi) + c=^ 0.

164. The equation (2) of the preceding article is the

same as equation (5) of Art. 161. If, therefore, the point

{xi, 2/i) be without the circle, as in the right-hand figure,

the polar is the same as the chord of contact of the real

tangents drawn through (x^^, y^).

If the point (x^, y^) be on the circle, the polar coincides

with the tangent at it. (Art. 150.)

GEOMETRICAL CONSTRUCTION FOR THE POLAR. 139

If the point (x^, y-f) be within the circle, then, as in

Art. 161, the equation (2) is the line joining the (imaginary)

points of contact of the two (imaginary) tangents that can

be drawn from (x-^ , y^.

We see therefore that the polar might have been

defined as follows :

The polar of a given point is the straight line which

passes through the (real or imaginary) points of contact of

tangents drawn from the given point ; also the pole of any

straight line is the point of intersection of tangents at the

points (real or imaginary) in which this straight line meets

the circle.

165. Geometrical construction for the imlar of a point.

The equation to OP^ which is the line joining (0, 0) to

^.e.

â– (!)â€¢

Also the polar of P is

xx^ + yVx â€” cC"

(2).

By Art. 69, the lines (1) and (2) are perpendicular to

one another. Hence OP is perpendicular to the polar

of P.

Also the length OP - ^x^+y},

140 COORDINATE GEOMETRY,

and the perpendicular, OiV, from upon (2)

Hence the product ON . OP â€” a^.

The polar of any point P is therefore constructed thus :

Join OP and on it (produced if necessary) take a point N

such that the rectangle ON . OP is equal to the square of

the radius of the circle.

Through N draw the straight line LL' perpendicular to

OP ; this is the polar required.

[It will be noted that the middle point N of any chord LL' lies on

the line joining the centre to the pole of the chord.]

166. To Jind the pole of a given line with respect to

any circle.

Let the equation to the given line be

^a; + % + (7-0 (1).

(1) Let the equation to the circle be

and let the required pole be {x-^, y-^.

Then (1) must be the equation to the polar of (oji, y^,

i.e. it is the same as the equation

xx-^ + yVi â€” cv^-O (2).

Comparing equations (1) and (2), Ave have

J B C '

so that Xt â€” â€” yz a^ and y, = â€” '^ a?.

C ^ C

The required pole is therefore the point

A , B ^

(2) Let the equation to the circle be

a;- + 2/2 + l.yx + 2fy + g=Q.

POLE AND POLAR. 141

If (a?!, 2/1) ^^ *^Â® required pole, then (1) must be

equivalent to the equation

xx^ +yyi + g{os + x,) +f{y + yi) + c=0, (Art. 1 63),

i.e. x(x^+g)+y{y^+f) + gxj^ +/2/i + c = (3).

Comparing (1) with (3), we therefore have

- sc^ + 9 _ Vi+f^ 9^1 +/2/1 + g

A " B C

By solving these equations we have the values of x^

and 2/1.

Ex. Find the pole of the straight line

9x+2j-28 = (1)

2vith respect to the circle

2x^+2y'^-Sx + 5ij-7 = (2).

If (.Tj, yj) be the required point the line (1) must coincide with the

polar of (x^, 2/i)' whose equation is

2xx^ + 2yy-^-i{x + x^)+^{y+y{j-l = 0,

i.e. x{4x^-d) + y{4:y^ + 5)-3xi + 5y^-U = (3).

Since (1) and (3) are the same, we have

4.'Ci - 3 _ 4|/i + 5 _ -Sxi + 5y^ - 14

9 r~ ~ ^28 â–

Hence Xi = 9yi + 12,

and 3.Ti-117i/i = 126.

Solving these equations we have Xj^ = S and t/^ = - 1, so that the

required point is (3, - 1).

167. If the 2')olar of a point P pass through a point T,

then the polar of T passes through P.

Let P and T be the points {x-^, y^ and (.Tg, 2/2) ^^-

spectively. (Fig. Art. 163.)

The polar of (x^, y^ with respect to the circle

a;^ + 2/2 _ ^2 jg

xx-^ + 2/2/1 = Cb^'

This straight line passes through the point T if

jcoa?! + 2/22/1 = ct^ â€¢â€¢â€¢ (!)â€¢

142 COORDINATE GEOMETRY.

Since the relation (1) is true it follows that the point

(x^, 2/1), i.e. P, lies on the straight line xx^-\-yy^ = (]?^ which

is the polar of {x^^ y^)^ i.e. T, with respect to the circle.

Hence the proposition.

Cor. The intersection, T, of the polars of two points,

P and Q, is the pole of the line PQ.

168. To find the length of the tangent that can he

drawn from the point (x^ , y-^ to the circles

(1) a?-^y'^ = a\

and (2) x? + y^ + 2gx + 2/^/ + c = 0.

If T be an external point (Fig. Art. 163), TQ a tangent

and the centre of the circle, then TQO is a right angle

and hence

(1) If the equation to the circle be x" ^-y"- â€” a^, is the

origin, OT'^ = x^- + y^.^ and OQ^ = a^.

Hence TQ" = x^^ + y^- - a\

(2) Let the equation to the circle be

x^ + y'^+ 2gx + 2fy + c = 0,

i.e. (x + gf-v{y+fY = f+f'-c.

In this case is the point (â€” g, â€”f) and

OQ^ â€” (radius)^ = g'^ +f^ â€” c.

Hence OT^ = [x, - (- g)Y + [y, - (-/)]' (Art. 20).

Therefore TQ^ ^ {x, + gf + (y, +ff - (f +p - c)

= <^ + Vi+'^g^i + 2/J/i + c.

In each case we see that (the equation to the circle

being written so that the coefficients of x^ and y" are each

unity) the square of the length of the tangent drawn to the

circle from the point {x^, y^) is obtained by substituting x^

and 2/1 for the current coordinates in the left-hand member

of the equation to the circle.

*169. To find the equation to the pair of tangents that

can he drawn from the point (x-^^ y^) to the circle x- + y^ â€” d^.

PAIR OF TANGENTS FROM ANY POINT. 143

Let (A, k) be any point on either of the tangents from

(^1, Vi)'

Since any straight line touches a circle if the perpen-

dicular on it from the centre is equal to the radius, the

perpendicular from the origin upon the line joining (x^, y^)

to (A, k) must be equal to a.

The equation to the straight line joining these two

points is

^ â€” Vi /

i.e. yQ^~~ ^'i) ~ ^ (^ ~ 2/1) + ^^1 ~~ ^^Ui â€” ^â–

__ kx^ â€” hy^

Hence , ^ = a,

J(h-x,f+(k-y,y

so that (kx^ â€” hy^y = a^ [{h â€” oo^y + {k â€” 2/1)^]-

Therefore the point (h, k) always lies on the locus

{x^y - xy^f = a" [{x - x^y + {y- y,y] (1).

This therefore is the required equation.

The equation (1) may be written in the form

= 2xyx^yi â€” 2a^xXj^ â€” 2a^yyi,

i.e. (x^ + y'^- CL^) {^i + y\ â€” Â«^) = x'^X{' + y^y^ + a^ + 2xyx{y^

â€” 2a^xXi â€” 2c^yy-^ â€” {xx^ + yy^ â€” a^y (2).

#170. In a later chapter we shall obtain the equation to the pair

of tangents to any curve of the second degree in a form analogous

to that of equation (2) of the previous article.

Similarly the equation to the pair of tangents that can be

drawn from {x-^, y-^ to the circle

If the equation to the circle be given in the form

a;2 + 2/2 + 2,9ra; + 2/?/ + c =

the equation to the tangents is, similarly,

{x' + ^2 + ^g^ + 2/y + c) {x^ + y^ + 2^a;i + 2jy^ + c)

= \xx^^yy^^g{x\x^^f{y^ry^) + cf (2).

144 COORDINATE GEOMETRY.

EXAMPLES. XIX.

Find the polar of the point

1. (1, 2) with respect to the circle a;- + y^ = 7.

2. (4, - 1) with respect to the circle 2x" + 2y^=ll,

3. (-2,3) with respect to the circle

x'^ + y^-4x-&y + 5 = 0.

4. (5, - I) with respect to the circle

Sx^ + Sy^-1x + 8y-9 = 0.

5. (a, - 6) with respect to the circle

x^ + y^ + 2ax - 2by + a^ - b- = 0.

Find the pole of the straight line

6. x + 2y = l with respect to the circle x^ + y^=5.

7. 2x-y = G with respect to the circle 5x^ + oy^ =9.

8. 2x + y + 12 = with respect to the circle

x^ + y^-4x + By-l=0.

9. 48;r - 54?/ + 53 = with respect to the circle

Sx^ + 3y^ + 5x-7y + 2 = 0.

10. ax + by + 3a^ + 36^ = with respect to the circle

x^ + y^ + 2ax + 2by = a^ + b^.

11. Tangents are drawn to the circle x^ + y^ = 12 at the points

where it is met by the circle x^ + y^ - 5x + Sy ~ 2 = ; find the point of

intersection of these tangents.

12. Find the equation to that chord of the circle x'^ + y^ = 81 which

is bisected at the point ( - 2, 3), and its pole with respect to the circle.

13. Prove that the polars of the point (1, - 2) with respect to the

circles whose equations are

x^ + y^ + Qy + 5 = and x'^ + y^ + 2x + 8y + 5 =

coincide ; prove also that there is another point the polars of which

with respect to these circles are the same and find its coordinates.

14. Find the condition that the chord of contact of tangents from

the point (x', y') to the circle x^+y^=a^ should subtend a right angle

at the centre.

15. Prove that the distances of two points, P and Q, each from

the polar of the other with respect to a circle, are to one another

inversely as the distances of the points from the centre of the circle.

16. Prove that the polar of a given point with respect to any one

of the circles x^-\-y'^-2kx-\-c'^ = 0, where k is variable, always passes

through a fixed point, whatever be the value of A\

[EXS. XIX.] POLAR EQUATION TO THE CIRCLE.

145

17. Tangents are drawn from the point {h, k) to the circle

x^ + y^ = aP; prove that the area of the triangle formed by them

and the straight line joining their points of contact is

ajJi^ + Ji^-aY

h^ + k^

Find the lengths of the tangents drawn

18. to the circle '2x^ + 2y'^â€”S from the point ( - 2, 3).

19. to the circle 3x^ + 3y^ -7x-6y = 12 from the point (6, - 7).

20. to the circle x^ + f' + Ihx - 36^ = from the point

(a + &, a-h).

21. Given the three circles

3ic2 + 37/-36a; + 81 = 0,

and a:2^.,y2_i6^_12y + 84 = 0,

find (1) the point from which the tangents to them are equal in

length, and (2) this length.

22. The distances from the origin of the centres of three circles

ic2-hy"^-2Xic = c^ (where c is a constant and X a variable) are in

geometrical progression ; prove that the lengths of the tangents drawn

to them from any point on the circle a;"^ + ^- = c-are also in geometrical

progression.

23. Find the equation to the pair of tangents drawn

(1) from the point (11, 3) to the. circle xr'-\-y^ â€” ^^,

(2) from the point (4, 5) to the circle

2x - + 2i/3 - 8a; + 12?/ + 21 r= 0.

171. To jincl the general equation of a circle referred

to polar coordinates.

Let be the origin, or pole, OX the initial line, G the

centre and a the radius of the

circle.

Let the polar coordinates of C

be R and a, so that 00 â€” M and

L XOC = a.

Let a radius vector through

at an angle 6 with the initial line

cut the circle in P and Q. Let

OP, or OQ, be r.

L.

10

146 COOKDINATE GEOMETRY.

Then {Trig. Art. 164) we have

CP- ^ OC'^ -r OP' -20C. OP cos C OF,

i.e. a''^P^ + r'-2Prcos(e~a),

i.e. r'- -2Rr cos {0 - a) + R"" - a" ^0 (1).

This is the required polar equation.

172. Particular cases of the general equation inpolar coordinates.

(1) Let the initial line be taken to go through the centre G. Then

a = 0, and the equation becomes

r2 - 2Rr cos ^ + E^ - a^ = 0.

(2) Let the pole be taken on the circle, so that

B=OG = a.

The general equation then becomes

^^M r^-2arcos{d-a) = 0,

i.e. r=2acos{d - a).

(3) Let the pole be on the circle and also let the initial line pass

through the centre of the circle. In this case

a=0, and R â€” a.

The general equation reduces then to the

simple form r = 2acos^.

This is at once evident from the figure. 0|

For, if OCA be a diameter, we have

0P=0^ cos ^,

i.e. r=2aQO8 0.

173. The equation (1) of Art. 171 is a quadratic

equation which, for any given value of 6, gives two

values of r. These two values in the figure are OP and

OQ.

If these two values be called r-^ and rg, we have, from

equation (1),

7\r^ = product of the roots â€” E^ â€” a"^,

i.e. OP.OQ^P'-a\

The value of the rectangle OP . OQ is therefore the

same for all values of 0. It follows that if we drew any

other line through to cut the circle in P^ and Q^ we

should have OP . OQ = OP^ .OQ^.

This is Euc. iii. 36, Cor.

POLAR EQUATION TO THE TANGENT. 147

174. Find the equation to the chord joining the points on the circle

r â€” 2a cos 6 whose vectorial angles are d^ and 6^, and deduce the equation

to the tangent at the point 6^.

The equation to any straight line in polar coordinates is (Art. 88)

^ = rcos((9-a) (1).

If this pass through the points (2a; cos ^j^, 6^ and (2asin^2Â» ^2)' ^^^

have

2a cos d-^ cos (^j-a)=^ = 2acos ^.3 cos {9.^~a) (2).

Hence cos (2^^ - a) + cos a = cos (2^., - a) + cos a,

i.e. 2^1 -a= -(2^0 -a),

since d^ and ^2 ^^^ iiotÂ» i'^ general, equal.

Hence a â€” d-^ + d.^^

and then, from (2), j? = 2a cos 6-^ cos d.-^.

On substitution in (1), the equation to the required chord is

r cos (^ - ^1 - ^2) = 2a cos ^1 cos ^2 (3).

The equation to the tangent at the point d^ is found, as in

Art. 150, by putting 6^ = 6^ in equation (3).

We thus obtain as the equation to the tangent

rcos (^-2^^) â€” 2acos2^j^. â€¢

As in the foregoing article it could be shewn that the equation to

the chord joining the points d^ and 6^ on the circle r= 2a cos {d - 7) is

r cos {d -9-^- 6., + 7] â€” 2a cos {9-^ - 7) cos (9.^ - 7)

and hence that the equation to the tangent at the point ^^ is

r cos (^ - 2^j + 7) = 2a cos2 (^j - 7).

EXAMPLES. XX.

1. Find the coordinates of the centre of the circle

r=A cos 9 + B sin 9.

2. Find the polar equation of a circle, the initial line being a

tangent. What does it become if the origin be on the circumference?

3. Draw the loci

(1) r=a; (2) r = a sin ^; (3) r = ttcos^; (4) r = asec^;

(5) r = acos(^-a); (6) r = asec(^-a).

4. Prove that the equations r = acos(^-a) and râ€”b sin (9 -a)

represent two circles which cut at right angles.

5. Prove that the equation r^ cos ^- a;* cos 2^ -2a- cos ^ =

represents a straight line and a circle.

10â€”2

148 COORDINATE GEOMETRY. [Exs. XX.]

6. Find the polar equation to the circle described on the straight

line joining the points {a, a) and (&, /3) as diameter.

7. Prove that the equation to the circle described on the straight

line joining the points (1, 60Â°) and (2, 30Â°) as diameter is

r^-r [cos {d - 60Â°) + 2 cos {0 - 30Â°)] + ^3 = 0.

8. Find the condition that the straight line

- = acos d + h sin^

r

may touch the circle r = 2c cos d.

175. To find the general equation to a circle referred to

oblique axes which meet at an angle to.

Let C be the centre and a the radius of the circle. Let

the coordinates of C be (h, k) so

that if CM, drawn parallel to the

axis of 2/, meets OX in M, then

OM=h and MC=.k.

Let P be any point on the

circle whose coordinates are x and

y. Draw PN, the ordinate of P,

and CL parallel to OX to meet

PNinL.

Then CL = MN = OiV - OM =- x - h,

and LP^NP-NL^NP -MG^y- k.

Also z CLP - z ONP= 180Â° - z PNX = 180Â° - tu.

Hence, since CL^- + LP^ - WL . LP cos CLP = a\

we have (x - h)2 + (y - k)2 + 2 (x - h) (y - k) cosco = 3?,

i.e. x^ + y^ + 2xy cos w - 2x [h + k cos w) ~2y ik + h cos w)

+ JiT + k^ + 2M COS w â€” or.

The required equation is therefore found.

176. As in Art. 142 it may be shewn that the

equation

aj- + ^xy cos oi-\- y" + '2gx + 2fy + c -

represents a circle and its radius and centre found.

OBLIQUE COORDINATES. J 49

Ex. If the axes he inclined at 60Â°, prove that the equation

x^ + xy+i/-^x-5y -2 â€” (1)

represents a circle and find its centre and radius.

If w be equal to 60Â°, so that cosw=|, the equation of Art. 175

D6COII1GS

x'^ + xy->ry - x{2h + Tc)-y{2k + h) + h'^ + l'^ + hh = a:^.

This equation agrees with (1) if

2/i + /c = 4 (2),

'ih + h^b (3),

and 7i2 + A;2 + 7i;e-a2= -2 (4).

Solving (2) and (3), we have h = l and k â€” 2. Equation (4) then

gives

- a2^7t2+F + ;iA; + 2z^9,

so that a = 3.

The equation (1) therefore represents a circle whose centre is the

point (1, 2) and whose radius is 3, the axes being inclined at 60Â°.

EXAMPLES. XXI.

Find the inclinations of the axes so that the following equations

may represent circles, and in each ease find the radius and centre ;

1. x'-xy-\-y'^~2gx-'ify = 0.

2. x'^ + fjZxy+if-^x-%y + 5 = Q.

3. The axes being inclined at an angle w, find the centre and

radius of the circle

a;2 + 2xy cos w + t/^ - Igx - 2fy = 0.

4. The axes being inclined at 45Â°, find the equation to the circle

whose centre is the point (2, 3) and whose radius is 4.

5. The axes being inclined at 60Â°, find the equation to the circle

whose centre is the point ( - 3, - 5) and whose radius is 6.

6. Prove that the equation to a circle whose radius is a and

which touches the axes of coordinates, which are inclined at an angle

to, is

x^ + 2xycoso} + y'^-2a (.T + ^)cot- +a^cot^-=0.

7. Prove that the straight line y â€” mx will touch the circle

x^ + 2xy cos (o + y^ + 2gx + 2fy + c =

if {g +fm)^ = c (1 + 2m cos o) + m^) .

8. The axes being inclined at an angle w, find the equation to the

circle whose diameter is the straight line joining the points

{x\ y') and {x", y").

150 COORDINATE GEOMETRY.

Coordinates of a point on a circle expressed in

terms of one single variable.

177. If, in the figure of Art. 139, we put the angle

MOP equal to a, the coordinates of the point P are easily

seen to be a cos a and a sin a.

These equations clearly satisfy equation (1) of that

article.

The position of the point P is therefore known when

the value of a is given, and it may be, for brevity, called

" the point a."

With the ordinary Cartesian coordinates we have to

give the values of two separate quantities x and y' (which

are however connected by the relation x' = Ja^ â€” y"^) to

express the position of a point P on the circle. The

above substitution therefore often simplifies solutions of

problems.

178. To find tlie equation to the straight line joining

two joints, a and y8, 07i the circle xr + y^ = a-.

Let the points be P and Q, and let OA^ be the perpen-

dicular from the origin on the straight line PQ ; then OJV

bisects the angle POQ, and hence

z X0]^= 1 ( z XOP + L XOQ) = HÂ« + ^)-

Also OK^ OP cos NOP = a cos Â°^^ .

The equation to PQ is therefore (Art. 53),

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