Copyright
S. L. (Sidney Luxton) Loney.

The elements of coordinate geometry online

. (page 9 of 26)
Online LibraryS. L. (Sidney Luxton) LoneyThe elements of coordinate geometry → online text (page 9 of 26)
Font size
QR-code for this ebook


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'^ — 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 — 0.

10. Find the condition that the straight line Ax-\-By + G — Q may
touch the circle

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

11. Find the equation to the tangent to the circle x - \-y'^ — 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 — x and y—-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 /• 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? ■\- y"^ — «', and let the
given point be (ajj, y^. [Fig. Art. 161.]

The equation to any tangent is, by Art. 155,
y — 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 — jiid\ — a Jl + m^,

i.e. y^ — 2rax-^y-^ + m^x^ = «- + ahn^^

i. e. m^ {xj^ — a^) — 2mx^y^ + y-^^ — 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)^ — 4 (.r/ — a"^) (2/1^ — a^) is positive, zero, or negative,

i. e. according as

a^ (— a"^ + Xj^ + y^) is positive, zero, or negative,

i. e. according as x-^ + y^ =■- 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^ <«^, 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"^ — a^ from the external point
(^1, 2/i)-



POLE AND POLAR.



137







•(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/' — a-^ (1),

and that at Q is

xx" + yy" —d^ (2).

Since these tangents pass through
T^ its coordinates {x-^ , y^ must satisfy
both (1) and (2).

Hence x-^ + y^y — 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^ — 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^ ■\-y^ — 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 ■\- yk — o?.

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

x-Ji + y^ — 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.






■(!)•





Also the polar of P is

xx^ + yVx — 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 — 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 — cv^-O (2).

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

J B C '

so that Xt — — yz a^ and y, = — '^ 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) ^^ *^® 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 ■

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^ ••• (!)•



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"- — 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 (— g, —f) and

OQ^ — (radius)^ = g'^ +f^ — 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^ — 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

^ — Vi /

i.e. yQ^~~ ^'i) ~ ^ (^ ~ 2/1) + ^^1 ~~ ^^Ui — ^■

__ kx^ — hy^

Hence , ^ = a,

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

so that (kx^ — hy^y = a^ [{h — oo^y + {k — 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 — 2a^xXj^ — 2a^yyi,
i.e. (x^ + y'^- CL^) {^i + y\ — «^) = x'^X{' + y^y^ + a^ + 2xyx{y^
— 2a^xXi — 2c^yy-^ — {xx^ + yy^ — 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'^—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^ — ^^,

(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 — 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 — 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 — E^ — 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 — 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^2» ^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 ^^^ iiot» i'^ general, equal.

Hence a — 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^^) — 2acos2^j^. •

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] — 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 r—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.

10—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, 60°) and (2, 30°) as diameter is

r^-r [cos {d - 60°) + 2 cos {0 - 30°)] + ^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= 180° - z PNX = 180° - 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 — 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 60°, prove that the equation

x^ + xy+i/-^x-5y -2 — (1)

represents a circle and find its centre and radius.

If w be equal to 60°, 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 — 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 60°.

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 45°, find the equation to the circle
whose centre is the point (2, 3) and whose radius is 4.

5. The axes being inclined at 60°, 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 — 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^ — 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) = H« + ^)-
Also OK^ OP cos NOP = a cos °^^ .

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



Online LibraryS. L. (Sidney Luxton) LoneyThe elements of coordinate geometry → online text (page 9 of 26)