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a + j8 . a+ i8 a â€” /?

X cos â€” p-^" + y sm â€” ^ - â€” a cos â€” - .

2 -^ 2 2

If we put P = a we have, as the equation to the tangent

at the point a,

X cos a + y sin a = a.

This may also be deduced from the equation of Art. 150

by putting x' = a cos a and y =cb sin a.

179. If the equation to the circle be in the more

general form

(x - hy + (y-ky^ a', (Art. 1 40),

THE CIRCLE. ONE VARIABLE. 151

we may express the coordinates of F in the form

(A + (X cos a, ^ + ct sin a).

For these values satisfy the above equation.

Here a is the angle LOP [Fig. Art. 140].

The equation to the straight line joining the points a and

^ can be easily shewn to be

/ IN Â« + yS / 7\ â€¢ a + )8 a. â€” ^

{x â€” h) cos â€” ^r^ -^ {y â€” Ic) sm â€” ^ â€” = a cos â€” ^ ,

and so the tangent at the point a is

{x â€” K) cos a + (2/ â€” li) sin a = a.

#180. Ex. Find the four common tangents to the two circles

ox'^ + 5i/ - 22x + iy + 20 = 0,

and 5a;2 + 5?/2 + 22a; -4y- 20 = 0.

The equations may be written

and (x + \^-f+{y-ir^ = SK

Any point on the first circle is (\^- + cos d, -| + sin d).

Any point on the second is ( - V- + 3 cos 0, f + 3 sin 0).

The equations to the tangents at these points are by the last

article

(a;- i^)cos^ + {y + |)sin^ = l (1),

and (a; + -V-)eos0 + (?/-|)sin0 = 3 (2).

These tangents coincide, in which case we have the common

tangents, if

cos ^ _ sin ^ _ - 11 cos ^ + 2 sin ^ - 5

cos <p ~ sin 9f> ~~ 11 cos 0-2 sin 0-15 ^ ^'

From the first pair of these equations we have (f) â€” dor(p = d + 180Â°.

If 0=:^, the second pair gives

-llcos^ + 2sin^-5

~ llcos^-2sin^-l"5 '

i.e. llcos^-2sin^ = 5 (4),

i.e. Il-2tan^ = 5^1 + tan2^.

On solving, we have tan ^ = | or - -^^.

152 COOEDINATE GEOMETRY.

Corresponding to tan^ = | we have the values sin ^ = 4, cos^ = f

which satisfy (4) ; we have also the values sin ^= - 1 and cos ^= - f ,

which do not satisfy it.

Corresponding to tan^= - y-> t^ie values which satisfy (4) are

sin^= -ff and cos^=/6.

If = 180Â°+ ^, the second pair of equations (3) give

- 11 cos ^ + 2 sin ^ - 5

~ -llcos^ + 2sin^-15'

i,e. 11 cos (9 -2 sin ^= -10 (5).

This gives (ll-2tan ^)2= 100(1 + tan2 6>),

so that tan^=-for^.

The corresponding values which satisfy (5) are found to be

sin 0=1, cos0=-| and sin^=-^t., cos0=-||.

The solutions of (3) are therefore

sin0=f, -i^, f, and -^^;

cos = 1, 2^5, -I, and -ff.

On substituting these values in equation (1), the common tangents

are found to be

3a; + 42/ = 10, 7.^-24?/ = 50,

4a;-3i/= 5, and 24.^+ 7?/ = 25.

181. We shall conclude this chapter with some mis-

cellaneous examples on loci.

Ex. I. Find the locus of a point P lohich moves so that its distance

from a given point is always in a given ratio [n : 1) to its distance

from another given point A.

Take as origin and the direction of OA as the axis of x. Let

the distance OA be rt, so that A is the point {a, 0).

If {Xy y) be the coordinates of any position of P we have

OP2=n2.^p2,

i.e. x'^-\-y'^ = n^[{x-a)^-\-y-'\,

i.e. (a;2 + 2/2)(n2-l)-2an2a; + 7i2a2=0 (1).

Hence, by Art. 143, the locus of P is a circle.

Let this circle meet the axis of x in the points C and D. Then OC

and OB are the roots of the equation obtained by putting y equal to

zero in (1).

-.-.- ^ ^ na , ^ ^ na

Hence 00= and 0D= 5.

n + 1 n-1

THE CIRCLE. EXAMPLES. 153

We therefore have

CA= and AD =

n+1 n-1

OG OB

^^^^^ GA=AD = ''-

The points C and D therefore divide the line OA in the given ratio,

and the required circle is on CD as diameter.

Ex. 2. From any 'point on one given circle tangents are drawn to

another given circle ; prove that the locus of the middle point of the

chord of contact is a third circle.

Take the centre of the first circle as origin and let the axis of x

pass through the centre of the second circle. Their equations are

then

a;2+2/2=a2 (1),

and (.'c-c)2 + r/2 = &2 (2),

where a and b are the radii, and c the distance between the centres, of

the circles.

Any point on (1) is {a cos 6, a sin 6) where is variable. Its chord

of contact with respect to (2) is

{x - c) {a cos 9 - c) -{-ya ^m 6 = 1^ (3) .

The middle point of this chord of contact is the point where it is

met by the perpendicular from the centre, viz. the point (c, 0).

The equation to this perpendicular is (Art. 70)

-{x-c)a sin 6 + [a cos 6 -c)y = (4).

Any equation deduced from (3) and (4) is satisfied by the coordi-

nates of the point under consideration. If we eliminate 6 from them,

we shall have an equation always satisfied by the coordinates of the

point, whatever be the value of 6. The result will thus be the equation

to the required locus.

Solving (3) and (4), we have

â€¢ /I ^^y

a Bin 6 =

and a cos 6 - c

7/ + (.r-6f'

_ y^ix-c)

~f+{^c^cf'

62 {x - c)

so that acos^ = c+ â€ž ,

y^ + {x- cy

Hence

a2 = a2 cos2 e + a^ sm^ e = c'^ + 2ch^ ^ ^~^ ^, + ^ .

y^+(x- c)- y^ + {x - c)2

The required locus is therefore

(a2 - c2) [?/2 + (.^ _ c)2] = 2c62 [x -c) + Â¥.

This is a circle and its centre and radius are easily found.

154 COORDINATE GEOMETRY.

Ex. 3. Find the locus of a point P which is such that its polar with

respect to one circle touches a second circle.

Taking the notation of the last article, the equations to the two

circles are

x^ + if=a- (1),

and (a?-c)2 + ^2^62 (2).

Let {h, Jc) he the coordinates of any position of P. Its polar with

respect to (1) is

xh + yk = a^ (3),

Also any tangent to (2) has its equation of the form (Art. 179)

[x - c) cos d + y sin d = b (4).

If then (3) be a tangent to (2) it must be of the form (4).

^, â€ž cos d sin 9 c cos d + b

Therefore â€” - â€” = â€” ^ â€” = ,5 .

h k a"^

These equations give

cos 6 {a^-ch) = hh, and sin d{a^-ch) = hk.

Squaring and adding, we have

(a2-c7i)2=62(/i2+^2) (5).

The locus of the point {h, k) is therefore the curve

h'^{x^ + y^) = {a^-cxf.

Aliter. The condition that (3) may touch (2) may be otherwise

found.

For, as in Art. 153, the straight line (3) meets the circle (2) in the

points whose abscissae are given by the equation

k^{x-c)^ + {a^-hxf^b%\

i. e. x^- {h^ + F) - 2x {ck^ + a'^h) + [k^c"^ + a" - b'^k^) = 0.

The line (3) will therefore touch (2) if

(cZ;2 + a^hf={h^ + F) (Fc^ + a^- b^k%

i.e. a h^h^ + B) = {ch-aY-,

which is equation (5).

Ex. 4. is a fixed point and P any point on a given circle ; OP

is joined and on it a point Q is taken so that OP . OQ = a constant

quantity k- ; prove that the locus of Q is a circle which becomes a

straight line ivhen lies on the original circle.

THE CIRCLE. EXAMPLES. 155

Let O be taken as pole and the line through the centre C as the

initial line. Let OC = d, and let the

radius of the circle be a. Qyf\

The equation to the circle is then

a2 = 7-2 + d^ - 2rd cos d, (Art. 171), Q \d C"

where OP=r and lPOG = d.

Let OQ be p, so that, by the given

condition, we have rp-=k' and hence ?- = â€” .

P

Substituting this value in the equation to the circle, we have

a2=^VtZ2-2â€” cos^ (1),

so that the equation to the locus of Q is

' -2 55 2^cos^=- (2.

But the equation to a circle, whose radius is a' and whose centre is

on the initial line at a distance d', is

r2-2rd'cos0 = a'2-d'2 (3).

Comparing (1) and (2), we see that the required locus is a circle,

such that

Z'2/7 hi

,r- and n"^ d'^-

Hence a 2= __ -^, ^^^^^.^ - 1 J = ^-^ .

The required locus is therefore a circle, of radius ~z^ ^ , whose

d^ - a"

kH

centre is on the same line as the original centre at a distance â€” ^ ^

'^ d - a^

from the fixed point.

When lies on the original circle the distance d is equal to a, and

the equation (1) becomes k'^ â€” 2drQ05dy i.e., in Cartesian coordinates,

_fc2

In this case the required locus is a straight line perpendicular

to OG.

When a second curve is obtained from a given curve by the above

geometrical process, the second curve is said to be the inverse of the

first curve and the fixed point O is called the centre of inversion.

The inverse of a circle is therefore a circle or a straight line

according as the centre of inversion is not, or is, on the circumference

of the original circle.

156 COORDINATE GEOMETRY.

Ex. 5. PQ is a straight line drawn through O, one of the common

points of two circles, and meets them'again in P and Q; find the locus of

the point S ivhich bisects the line PQ.

Take O as the origin, let the radii of the two circles be R and R',

and let the lines joining their centres to O make angles a and a' with

the initial line.

The equations to the two circles are therefore, {Art. 172 (2)},

r=2i?cos{^-a), and r = 2R' cos{d- a').

Hence, if S be the middle point of PQ, we have

20S =0P+0Q = 2R cos {d-a) + 2R' cos {6- a').

The locus of the point S is therefore

r = Rcos{d-a) + R'co3{e-a')

= {R cos a + R' cos a') cos ^ + (JR sin a + R' sin a'} sin 6

= 2R''co8{d-a") (1),

where 2R" cos a" = R cos a + R' cos a',

and 2R" sin a" =Rsma + R' sin a'.

Hence R" = i s/R^ + R'^ + 2RR' cos (a - a') ,

R sin a + R' sin a'

and tana"=

jR cos a + jR' cos a' '

From (1) the locus of S is a circle, whose radius is JR", which

passes through the origin and is such that the line joining O to its

centre is inclined at an angle a" to the initial line.

EXAMPLES. XXII.

1. A point moves so that the sum of the squares of its distances

from the four sides of a square is constant ; prove that it always lies

on a circle,

2. A point moves so that the sum of the squares of the perpendi-

culars let fall from it on the sides of an equilateral triangle is constant;

prove that its locus is a circle.

3. A point moves so that the sum of the squares of its distances

from the angular points of a triangle is constant ; prove that its locus

is a circle.

4. Find the locus of a point which moves so that the square of

the tangent drawn from it to the circle x^ + y^=a^ is equal to c times

its distance from the straight line lx + my + n = 0.

5. Find the locus of a point whose distance from a fixed point is

in a constant ratio to the tangent drawn from it to a given circle.

[EXS. XXII.] EXAMPLES. 157

6. Find the locus of the vertex of a triangle, given (1) its base and

the sum of the squares of its sides, (2) its base and the sum of m times

the square of one side and n times the square of the other.

7. A point moves so that the sum of the squares of its distances

from n fixed points is given. Prove that its locus is a circle.

8. Whatever be the value of a, prove that the locus of the inter-

section of the straight lines

X cos a + y sin a â€” a and x sin a-y cos a â€” b

is a circle.

9. From a point P on a circle perpendiculars P3I and FN are

drawn to two radii of the cu'cle which are not at right angles ; find

the locus of the middle point of 3IN.

10. Tangents are drawn to a circle from a point which always

lies on a given line ; prove that the locus of the middle point of the

chord of contact is another circle.

11. Find the locus of the middle points of chords of the circle

x^ + y^ = aP which pass through the fixed point {h, k).

12. Find the locus of the middle points of chords of the circle

x'^ + y^=a? which subtend a right angle at the point (c, 0).

13. is a fixed point and P any point on a fixed circle ; on OP

is taken a point Q such that OQ is in a constant ratio to OP ; prove

that the locus of Q is a circle.

14. is a fixed point and P any point on a given straight line ;

OP is joined and on it is taken a point Q such that OP . OQ = k-;

prove that the locus of Q, i. e. the inverse of the given straight line

with respect to 0, is a circle which passes through O.

15. One vertex of a triangle of given species is fixed, and another

moves along the circumference of a fixed circle ; prove that the locus

of the remaining vertex is a circle and find its radius.

16. O is any point in the plane of a circle, and OP^P^ any chord

of the circle which passes through O and meets the circle in Pj and

P. 2. On this chord is taken a point Q such that OQ is equal to (1) the

arithmetic, (2) the geometric, and (3) the harmonic mean between OP^

and 0P<^\ in each case find the equation to the locus of Q.

17. Find the locus of the point of intersection of the tangent to

any circle and the perpendicular let fall on this tangent from a fixed

point on the circle.

18. A circle touches the axis of x and cuts off a constant length

21 from the axis of y ; prove that the equation of the locus of its centre

is 2/^ - x^ = l^ cosec^ w, the axes being inclined at an angle w.

158 COORDINATE GEOMETHY. [ExS.

19. A straight line moves so that the product of the jDerpendi-

culars on it from two fixed points is constant. Prove that the locus

of the feet of the perpendiculars from each of these points upon the

straight line is a circle, the same for each.

20. is a fixed point and AP and BQ are two fixed parallel

straight lines ; BOA is perpendicular to both and POQ is a right

angle. Prove that the locus of the foot of the perpendicular drawn

from O upon PQ is the circle on AB as diameter.

21. Two rods, of lengths a and b, slide along the axes, which are

rectangular, in such a manner that their ends are always concyclic ;

prove that the locus of the centre of the circle passing through these

ends is the curve 4 {x^ - y'^) â€” a'^ - b^.

22. Shew that the locus of a point, which is such that the

tangents from it to two given concentric circles are inversely as the

radii, is a concentric circle, the square of whose radius is equal to the

sum of the squares of the radii of the given circles.

23. Shew that if the length of the tangent from a point P to the

circle x^ + y^=a-^ be four times the length of the tangent from it to the

circle {x - a)- + y^ = a?, then P lies on the circle

Ihx'^ + 15?/2 - ^lax + a2 = 0.

Prove also that these three circles pass through two points and that

the distance between the centres of the first and third circles is

sixteen times the distance between the centres of the second and

third circles.

24. Find the locus of the foot of the perpendicular let fall from

the origin upon any chord of the circle x- + y'^-\-2gx + 2fy + c=i(i which

subtends a right angle at the origin.

Find also the locus of the middle points of these chords.

25. Through a fixed point O are drawn two straight lines OPQ

and ORS to meet the circle in P and Q, and jR and S, respectively.

Prove that the locus of the point of intersection of PS and QR, as also

that of the point of intersection of PR and QS, is the polar of with

respect to the circle.

26. ^) -B, C, and D are four points in a straight line; prove that

the locus of a point P, such that the angles APB and CPD are equal,

is a circle.

27. The polar of P with respect to the circle x'^-\-y'^ â€” a- touches

the circle {x - of + {y - ^)"â€”b^ ; prove that its locus is the curve given

by the equation {ax + ^y-a^f â€” b^ {x^ + y^) .

28. A tangent is drawn to the circle {x - a)" + y" = b- and a perpen-

dicular tangent to the circle {x + a)'^ + y- = c" ; find the locus of their

point of intersection, and prove that the bisector of the angle between

them always touches one or other of two fixed circles.

XXII.] EXAMPLES. ' 159

29. Ill any circle prove that the perpendicular from any point of

it on the line joining the points of contact of two tangents is a mean

proportional between the perpendiculars from the point upon the two

tangents.

30. From any point on the circle

a;2 + if + 2gx + 2fy + c^0

tangents are drawn to the circle

x^ + ij'^ + 2gx + 2fy + c sin^ a + {g'^+f^) cos^ a = 0;

prove that the angle between them is 2a.

31. The angular points of a triangle are the points

(a cos a, a sin a), (acos/S, asin/3), and (a cos 7, a sin 7);

prove that the coordinates of the orthocentre of the triangle are

a (cos a + cos /3 + cos 7) and a (sin a + sin /3 + sin 7).

Hence prove that if ^i, B, C, and D be four points on a circle the

orthocentres of the four triangles ABC, BCD, CDA, and DAB lie on

a circle.

32. A variable circle passes through the point of intersection

of any two straight lines and cuts ofi from them portions OP and OQ

such that m . OP + n . OQ is equal to unity ; prove that this circle

always passes through a fixed point.

33. Find the length of the common chord of the circles, whose

equations are (x-af + y'^ = a? and x^ + (1/ - h)-=^JP, and prove that the

equation to the circle whose diameter is this common chord is

(a2 + &2) (a;2 + yi-^ ^2ab {bx + aij) .

34. Prove that the length of the common chord of the two circles

whose equations are

(a;-a)2 + (?/-fc)2 = c-2 and {x-hf + iy-af^c^

is V^c- - 2 (a - h)-K

Hence find the condition that the two circles may touch.

35. Find the length of the common chord of the circles

x^ + y^-2ax-4ay -4^0^ = and X' + y'-oax + 4:ay = 0.

Find also the equations of the common tangents and shew that

the length of each is 4a.

36. Find the equations to the common tangents of the circles

(1) x'^ + 7f-2x~&y + 9 = and x- + y-^ + ex-2y + 1 = 0,

(2) x^ + 7/2 = 6-2 and {x - a)^ + y^ = h-.

CHAPTER IX.

SYSTEMS OF CIRCLES.

[This chapter may he omitted hy the student on a first

reading of the subject.]

182. Orthogonal Circles.

said to intersect orthogonally when

the tangents at their points of

intersection are at right angles.

If the two circles intersect at

P, the radii O^P and O.^P., which

are perpendicular to the tangents

at P, must also be at right angles.

Def. Two circles are

Hence

O^Oi=O^P^-vO.JP\

i.e. the square of the distance between the centres must be

equal to the sum of the squares of the radii.

Also the tangent from Oo, to the other circle is equal to

the radius a^,, i.e. if two circles be orthogonal the length of

the tangent drawn from the centre of one circle to the

second circle is equal to the radius of the first.

Either of these two conditions will determine whether

the circles are orthogonal.

The centres of the circles

a;2 + 2/2 + 2f;^ + 2/?/ + c = and a;- + ?/2 + 2^'a; + 2/'y + c' = 0,

are the points {-g, -/) and (-^', -/') ; also the squares of their

radii are g^+f~- c and g'^ +f'^ - c'.

RADICAL AXIS OF TWO CIRCLES. 161

They therefore cut orthogonally if

i.e. if 2gg' + 2ff=c + c'.

1S3. Radical Axis. Def. The radical axis of

two circles is the locus of a point which moves so that the

lengths of the tangents drawn from it to the two circles are

equal.

Let the equations to the circles be

x' + y' + 2gx + 2fi/ + G^0 (1),

and x" + i/^ + 2gjX + 2fy + c^ = (2),

and let (x^, y^ be any point such that the tangents from it

to these circles are equal.

By Art. 168, we have

^1^ + Vx + ^9^1 + 2/2/1 + c = a^i" + 2/r + 2g^x^ + 2fy^ + c^,

i.e. 2x^{g-g,) + 2y,{f-f,) + G-C:, = 0.

But this is the condition that the point (iCj, 2/1) should

lie on the locus

2x{g-g,) + 2y{f-A) + c-c, = (3).

This is therefore the equation to the radical axis, and it

is clearly a straight line.

It is easily seen that the radical axis is perpendicular

to the line joining the centres of the circles. For these

centres are the points (â€” g, -f) and (-^1, â€” /i)- The

"m" of the line ioining them is therefore -^ â€” 7 {,

^ ^ -9x-{-9)

i.e. -tA.

9-9i

The "?7i" of the line (3) is - 7â€” Â§.

The product of these two " m's " is - 1 .

Hence, by Art. 69, the radical axis and the line joining

the centres are perpendicular.

L. 11

162

COORDINATE GEOMETRY.

184. A geometrical construction can be given

for the radical axis of two circles.

R,

^^^

\ >v

~~^^^/

^ ^v

/ \

A.

^ â€”

Ss

/ *

/ \

/^

7\

1

I

' \

o.

o

o.

Fig. 1.

Fig. 2.

If the circles intersect in real points, F and Q^ as in

Fig. 1, the radical axis is clearly the straight line PQ.

For if T be any point on PQ and TR and TS be the

tangents from it to the circles we have, by Euc. iii. 36,

TR'

TS\

TP . TQ

If they do not intersect in real points, as in the second

figure, let their radii be % and a^, and let 2^ be a point such

that the tangents TR and TS are equal in length.

Draw TO perpendicular to 0^0.2-

Since TR^=TS\

we have T0{- - O^R'' = TOi - O^S^

TO'' + 0^0'' - a^- - TO'' + OO.J - ai,

0^0''-00i=^a^-ai,

{0^0 - 00^) (0^0 + 00.) = ai' - a^\

0^0 â€” OOo =""7777^ = ^ constant quantity.

UiU.2

Hence is a fixed point, since it divides the fixed

straight line O^O^ into parts whose difference is constant.

Therefore, since O^OT is a right angle, the locus of T,

i.e. the radical axis, is a fixed straight line perpendicular to

the line joining the centres.

t.e.

%.e.

%.e,

I.e.

RADICAL AXIS. 163

185. If the equations to the circles in Art. 183 be

written in the form aS'=0 and aS" = 0, the equation (3) to

the radical axis may be written S â€” S' = 0, and therefore

the radical axis passes through the common points, real or

imaginary, of the circles S = and aS" = 0.

In the last article we saw that this was true geometri-

cally for the case in which the circles meet in real points.

When the circles do not geometrically intersect, as in

Fig. 2, we must then look upon the straight line TO as

passing through the imaginary points of intersection of the

two circles.

186. The radical axes of tJwee circles^ taken in 2)cdrs^

meet iri a j^oint.

Let the equations to the three circles be

^ = (1),

'S^'^O (2),

and S"=^0 (3).

The radical axis of the circles (1) and (2) is the straight

line

S->S' = (4).

The radical axis of (2) and (3) is the straight line

.S"-.S"'=:0 (5).

If we add equation (5) to equation (4) we shall have the

equation of a straight line through their points of inter-

section.

Hence .S'-aS'"-0 (6)

is a straight line through the intersection of (4) and (5).

But (6) is the radical axis of the circles (3) and (1).

Hence the three radical axes of the three circles, taken

in pairs, meet in a point.

This point is called the Radical Centre of the three

circles.

This may also be easily proved geometrically. For let

the three circles be called A, B, and C, and let the radical

axis of A and B and that of B and C meet in a point 0.

11â€”2

164 COORDINATE GEOMETKY.

By the definition of the radical axis, the tangent from

to the circle A = the tansrent from ^ - -^

to the circle B, and the tangent f V j

from to the circle B = tangent z'^' \ ^^\^^

from it to the circle C. f \ Ip

Hence the tangent from to >v ^/^v^' /^"^^

the circle A = the tangent from it ^^7\ \ \

to the circle C, i.e. is also a I ^^ j-

point on the radical axis of the V J

circles A and C. â€”

187. If S=0 and S' = he the equations of two circles,

the equation of any circle through their 2>oints of inter-

section is S â€” \S'. Also the equation to any circle, such that

the radical axis of it and Sâ€”0 is u = 0, is S + \u ~ 0.

For wherever S =0 and >S" = are both satisfied the

equation S = XS' is clearly satisfied, so that S = \S' is some

locus through the intersections of aS' = and *S"= 0.

Also in both S and S' the coefiicients of x^ and y'^ are

equal and the coefiicient of xy is zero. The same statement

is therefore true for the equation S=XS'. Hence the

proposition.

Again, since u is only of the first degree, therefore in

S + Xu the coefficients of ay^ and y^ are equal and the

coefficient of xy is zero, so that S + Xu = is clearly a circle.

Also it passes through the intersections of *S^ = and u â€” 0.

EXAMPLES. XXIII.

Prove that the following pairs of circles intersect orthogonally :

1. x'^ + y^-2ax + c = and x^ + y^ + 2by -c = 0.

2. x^ + y- - 2ax + 2hy + c = and x^ + y^ + 2hx + 2ay -c = 0.

3. Find the equation to the circle which passes through the origin

and cuts orthogonally each of the circles

x^ + y^-ex + 8 = and x^ + y^-2x-2y = 7.

Find the radical axis of the pairs of circles

4. ic2 + 2/2=144 and x^ + y^-15x + lly = 0.

5. x^ + y^-3x-4:y + 5=0 and Sx^- + By^-7x + 8y + ll=0.

RADICAL AXIS. EXAMPLES. 165

6. x^ + y - xy + Qx-ly + 8 = &nd x^ + ij' -xy-4 = 0,

the axes being inclined at 120Â°.

Find the radical centre of the sets of circles

7. x^+y^ + x + 2y + S = 0, x^ + 2f-h2x + 4:y + 5 = 0,

and x^ + y^-7x-8y-d = 0.

8. (a;-2)2+(z/-3)2 = 36, (re + 3)2 +(7/ + 2)2 = 49,

and (a;-4)2 + (i/ + 5)2=64.

9. Prove that the square of the tangent that can be drawn from

any point on one circle to another circle is equal to twice the product

of the perpendicular distance of the point from the radical axis of the

two circles, and the distance between their centres.

10. Prove that a common tangent to two circles is bisected by the

radical axis.

11. Find the general equation of aU circles any pair of which have

X cos â€” p-^" + y sm â€” ^ - â€” a cos â€” - .

2 -^ 2 2

If we put P = a we have, as the equation to the tangent

at the point a,

X cos a + y sin a = a.

This may also be deduced from the equation of Art. 150

by putting x' = a cos a and y =cb sin a.

179. If the equation to the circle be in the more

general form

(x - hy + (y-ky^ a', (Art. 1 40),

THE CIRCLE. ONE VARIABLE. 151

we may express the coordinates of F in the form

(A + (X cos a, ^ + ct sin a).

For these values satisfy the above equation.

Here a is the angle LOP [Fig. Art. 140].

The equation to the straight line joining the points a and

^ can be easily shewn to be

/ IN Â« + yS / 7\ â€¢ a + )8 a. â€” ^

{x â€” h) cos â€” ^r^ -^ {y â€” Ic) sm â€” ^ â€” = a cos â€” ^ ,

and so the tangent at the point a is

{x â€” K) cos a + (2/ â€” li) sin a = a.

#180. Ex. Find the four common tangents to the two circles

ox'^ + 5i/ - 22x + iy + 20 = 0,

and 5a;2 + 5?/2 + 22a; -4y- 20 = 0.

The equations may be written

and (x + \^-f+{y-ir^ = SK

Any point on the first circle is (\^- + cos d, -| + sin d).

Any point on the second is ( - V- + 3 cos 0, f + 3 sin 0).

The equations to the tangents at these points are by the last

article

(a;- i^)cos^ + {y + |)sin^ = l (1),

and (a; + -V-)eos0 + (?/-|)sin0 = 3 (2).

These tangents coincide, in which case we have the common

tangents, if

cos ^ _ sin ^ _ - 11 cos ^ + 2 sin ^ - 5

cos <p ~ sin 9f> ~~ 11 cos 0-2 sin 0-15 ^ ^'

From the first pair of these equations we have (f) â€” dor(p = d + 180Â°.

If 0=:^, the second pair gives

-llcos^ + 2sin^-5

~ llcos^-2sin^-l"5 '

i.e. llcos^-2sin^ = 5 (4),

i.e. Il-2tan^ = 5^1 + tan2^.

On solving, we have tan ^ = | or - -^^.

152 COOEDINATE GEOMETRY.

Corresponding to tan^ = | we have the values sin ^ = 4, cos^ = f

which satisfy (4) ; we have also the values sin ^= - 1 and cos ^= - f ,

which do not satisfy it.

Corresponding to tan^= - y-> t^ie values which satisfy (4) are

sin^= -ff and cos^=/6.

If = 180Â°+ ^, the second pair of equations (3) give

- 11 cos ^ + 2 sin ^ - 5

~ -llcos^ + 2sin^-15'

i,e. 11 cos (9 -2 sin ^= -10 (5).

This gives (ll-2tan ^)2= 100(1 + tan2 6>),

so that tan^=-for^.

The corresponding values which satisfy (5) are found to be

sin 0=1, cos0=-| and sin^=-^t., cos0=-||.

The solutions of (3) are therefore

sin0=f, -i^, f, and -^^;

cos = 1, 2^5, -I, and -ff.

On substituting these values in equation (1), the common tangents

are found to be

3a; + 42/ = 10, 7.^-24?/ = 50,

4a;-3i/= 5, and 24.^+ 7?/ = 25.

181. We shall conclude this chapter with some mis-

cellaneous examples on loci.

Ex. I. Find the locus of a point P lohich moves so that its distance

from a given point is always in a given ratio [n : 1) to its distance

from another given point A.

Take as origin and the direction of OA as the axis of x. Let

the distance OA be rt, so that A is the point {a, 0).

If {Xy y) be the coordinates of any position of P we have

OP2=n2.^p2,

i.e. x'^-\-y'^ = n^[{x-a)^-\-y-'\,

i.e. (a;2 + 2/2)(n2-l)-2an2a; + 7i2a2=0 (1).

Hence, by Art. 143, the locus of P is a circle.

Let this circle meet the axis of x in the points C and D. Then OC

and OB are the roots of the equation obtained by putting y equal to

zero in (1).

-.-.- ^ ^ na , ^ ^ na

Hence 00= and 0D= 5.

n + 1 n-1

THE CIRCLE. EXAMPLES. 153

We therefore have

CA= and AD =

n+1 n-1

OG OB

^^^^^ GA=AD = ''-

The points C and D therefore divide the line OA in the given ratio,

and the required circle is on CD as diameter.

Ex. 2. From any 'point on one given circle tangents are drawn to

another given circle ; prove that the locus of the middle point of the

chord of contact is a third circle.

Take the centre of the first circle as origin and let the axis of x

pass through the centre of the second circle. Their equations are

then

a;2+2/2=a2 (1),

and (.'c-c)2 + r/2 = &2 (2),

where a and b are the radii, and c the distance between the centres, of

the circles.

Any point on (1) is {a cos 6, a sin 6) where is variable. Its chord

of contact with respect to (2) is

{x - c) {a cos 9 - c) -{-ya ^m 6 = 1^ (3) .

The middle point of this chord of contact is the point where it is

met by the perpendicular from the centre, viz. the point (c, 0).

The equation to this perpendicular is (Art. 70)

-{x-c)a sin 6 + [a cos 6 -c)y = (4).

Any equation deduced from (3) and (4) is satisfied by the coordi-

nates of the point under consideration. If we eliminate 6 from them,

we shall have an equation always satisfied by the coordinates of the

point, whatever be the value of 6. The result will thus be the equation

to the required locus.

Solving (3) and (4), we have

â€¢ /I ^^y

a Bin 6 =

and a cos 6 - c

7/ + (.r-6f'

_ y^ix-c)

~f+{^c^cf'

62 {x - c)

so that acos^ = c+ â€ž ,

y^ + {x- cy

Hence

a2 = a2 cos2 e + a^ sm^ e = c'^ + 2ch^ ^ ^~^ ^, + ^ .

y^+(x- c)- y^ + {x - c)2

The required locus is therefore

(a2 - c2) [?/2 + (.^ _ c)2] = 2c62 [x -c) + Â¥.

This is a circle and its centre and radius are easily found.

154 COORDINATE GEOMETRY.

Ex. 3. Find the locus of a point P which is such that its polar with

respect to one circle touches a second circle.

Taking the notation of the last article, the equations to the two

circles are

x^ + if=a- (1),

and (a?-c)2 + ^2^62 (2).

Let {h, Jc) he the coordinates of any position of P. Its polar with

respect to (1) is

xh + yk = a^ (3),

Also any tangent to (2) has its equation of the form (Art. 179)

[x - c) cos d + y sin d = b (4).

If then (3) be a tangent to (2) it must be of the form (4).

^, â€ž cos d sin 9 c cos d + b

Therefore â€” - â€” = â€” ^ â€” = ,5 .

h k a"^

These equations give

cos 6 {a^-ch) = hh, and sin d{a^-ch) = hk.

Squaring and adding, we have

(a2-c7i)2=62(/i2+^2) (5).

The locus of the point {h, k) is therefore the curve

h'^{x^ + y^) = {a^-cxf.

Aliter. The condition that (3) may touch (2) may be otherwise

found.

For, as in Art. 153, the straight line (3) meets the circle (2) in the

points whose abscissae are given by the equation

k^{x-c)^ + {a^-hxf^b%\

i. e. x^- {h^ + F) - 2x {ck^ + a'^h) + [k^c"^ + a" - b'^k^) = 0.

The line (3) will therefore touch (2) if

(cZ;2 + a^hf={h^ + F) (Fc^ + a^- b^k%

i.e. a h^h^ + B) = {ch-aY-,

which is equation (5).

Ex. 4. is a fixed point and P any point on a given circle ; OP

is joined and on it a point Q is taken so that OP . OQ = a constant

quantity k- ; prove that the locus of Q is a circle which becomes a

straight line ivhen lies on the original circle.

THE CIRCLE. EXAMPLES. 155

Let O be taken as pole and the line through the centre C as the

initial line. Let OC = d, and let the

radius of the circle be a. Qyf\

The equation to the circle is then

a2 = 7-2 + d^ - 2rd cos d, (Art. 171), Q \d C"

where OP=r and lPOG = d.

Let OQ be p, so that, by the given

condition, we have rp-=k' and hence ?- = â€” .

P

Substituting this value in the equation to the circle, we have

a2=^VtZ2-2â€” cos^ (1),

so that the equation to the locus of Q is

' -2 55 2^cos^=- (2.

But the equation to a circle, whose radius is a' and whose centre is

on the initial line at a distance d', is

r2-2rd'cos0 = a'2-d'2 (3).

Comparing (1) and (2), we see that the required locus is a circle,

such that

Z'2/7 hi

,r- and n"^ d'^-

Hence a 2= __ -^, ^^^^^.^ - 1 J = ^-^ .

The required locus is therefore a circle, of radius ~z^ ^ , whose

d^ - a"

kH

centre is on the same line as the original centre at a distance â€” ^ ^

'^ d - a^

from the fixed point.

When lies on the original circle the distance d is equal to a, and

the equation (1) becomes k'^ â€” 2drQ05dy i.e., in Cartesian coordinates,

_fc2

In this case the required locus is a straight line perpendicular

to OG.

When a second curve is obtained from a given curve by the above

geometrical process, the second curve is said to be the inverse of the

first curve and the fixed point O is called the centre of inversion.

The inverse of a circle is therefore a circle or a straight line

according as the centre of inversion is not, or is, on the circumference

of the original circle.

156 COORDINATE GEOMETRY.

Ex. 5. PQ is a straight line drawn through O, one of the common

points of two circles, and meets them'again in P and Q; find the locus of

the point S ivhich bisects the line PQ.

Take O as the origin, let the radii of the two circles be R and R',

and let the lines joining their centres to O make angles a and a' with

the initial line.

The equations to the two circles are therefore, {Art. 172 (2)},

r=2i?cos{^-a), and r = 2R' cos{d- a').

Hence, if S be the middle point of PQ, we have

20S =0P+0Q = 2R cos {d-a) + 2R' cos {6- a').

The locus of the point S is therefore

r = Rcos{d-a) + R'co3{e-a')

= {R cos a + R' cos a') cos ^ + (JR sin a + R' sin a'} sin 6

= 2R''co8{d-a") (1),

where 2R" cos a" = R cos a + R' cos a',

and 2R" sin a" =Rsma + R' sin a'.

Hence R" = i s/R^ + R'^ + 2RR' cos (a - a') ,

R sin a + R' sin a'

and tana"=

jR cos a + jR' cos a' '

From (1) the locus of S is a circle, whose radius is JR", which

passes through the origin and is such that the line joining O to its

centre is inclined at an angle a" to the initial line.

EXAMPLES. XXII.

1. A point moves so that the sum of the squares of its distances

from the four sides of a square is constant ; prove that it always lies

on a circle,

2. A point moves so that the sum of the squares of the perpendi-

culars let fall from it on the sides of an equilateral triangle is constant;

prove that its locus is a circle.

3. A point moves so that the sum of the squares of its distances

from the angular points of a triangle is constant ; prove that its locus

is a circle.

4. Find the locus of a point which moves so that the square of

the tangent drawn from it to the circle x^ + y^=a^ is equal to c times

its distance from the straight line lx + my + n = 0.

5. Find the locus of a point whose distance from a fixed point is

in a constant ratio to the tangent drawn from it to a given circle.

[EXS. XXII.] EXAMPLES. 157

6. Find the locus of the vertex of a triangle, given (1) its base and

the sum of the squares of its sides, (2) its base and the sum of m times

the square of one side and n times the square of the other.

7. A point moves so that the sum of the squares of its distances

from n fixed points is given. Prove that its locus is a circle.

8. Whatever be the value of a, prove that the locus of the inter-

section of the straight lines

X cos a + y sin a â€” a and x sin a-y cos a â€” b

is a circle.

9. From a point P on a circle perpendiculars P3I and FN are

drawn to two radii of the cu'cle which are not at right angles ; find

the locus of the middle point of 3IN.

10. Tangents are drawn to a circle from a point which always

lies on a given line ; prove that the locus of the middle point of the

chord of contact is another circle.

11. Find the locus of the middle points of chords of the circle

x^ + y^ = aP which pass through the fixed point {h, k).

12. Find the locus of the middle points of chords of the circle

x'^ + y^=a? which subtend a right angle at the point (c, 0).

13. is a fixed point and P any point on a fixed circle ; on OP

is taken a point Q such that OQ is in a constant ratio to OP ; prove

that the locus of Q is a circle.

14. is a fixed point and P any point on a given straight line ;

OP is joined and on it is taken a point Q such that OP . OQ = k-;

prove that the locus of Q, i. e. the inverse of the given straight line

with respect to 0, is a circle which passes through O.

15. One vertex of a triangle of given species is fixed, and another

moves along the circumference of a fixed circle ; prove that the locus

of the remaining vertex is a circle and find its radius.

16. O is any point in the plane of a circle, and OP^P^ any chord

of the circle which passes through O and meets the circle in Pj and

P. 2. On this chord is taken a point Q such that OQ is equal to (1) the

arithmetic, (2) the geometric, and (3) the harmonic mean between OP^

and 0P<^\ in each case find the equation to the locus of Q.

17. Find the locus of the point of intersection of the tangent to

any circle and the perpendicular let fall on this tangent from a fixed

point on the circle.

18. A circle touches the axis of x and cuts off a constant length

21 from the axis of y ; prove that the equation of the locus of its centre

is 2/^ - x^ = l^ cosec^ w, the axes being inclined at an angle w.

158 COORDINATE GEOMETHY. [ExS.

19. A straight line moves so that the product of the jDerpendi-

culars on it from two fixed points is constant. Prove that the locus

of the feet of the perpendiculars from each of these points upon the

straight line is a circle, the same for each.

20. is a fixed point and AP and BQ are two fixed parallel

straight lines ; BOA is perpendicular to both and POQ is a right

angle. Prove that the locus of the foot of the perpendicular drawn

from O upon PQ is the circle on AB as diameter.

21. Two rods, of lengths a and b, slide along the axes, which are

rectangular, in such a manner that their ends are always concyclic ;

prove that the locus of the centre of the circle passing through these

ends is the curve 4 {x^ - y'^) â€” a'^ - b^.

22. Shew that the locus of a point, which is such that the

tangents from it to two given concentric circles are inversely as the

radii, is a concentric circle, the square of whose radius is equal to the

sum of the squares of the radii of the given circles.

23. Shew that if the length of the tangent from a point P to the

circle x^ + y^=a-^ be four times the length of the tangent from it to the

circle {x - a)- + y^ = a?, then P lies on the circle

Ihx'^ + 15?/2 - ^lax + a2 = 0.

Prove also that these three circles pass through two points and that

the distance between the centres of the first and third circles is

sixteen times the distance between the centres of the second and

third circles.

24. Find the locus of the foot of the perpendicular let fall from

the origin upon any chord of the circle x- + y'^-\-2gx + 2fy + c=i(i which

subtends a right angle at the origin.

Find also the locus of the middle points of these chords.

25. Through a fixed point O are drawn two straight lines OPQ

and ORS to meet the circle in P and Q, and jR and S, respectively.

Prove that the locus of the point of intersection of PS and QR, as also

that of the point of intersection of PR and QS, is the polar of with

respect to the circle.

26. ^) -B, C, and D are four points in a straight line; prove that

the locus of a point P, such that the angles APB and CPD are equal,

is a circle.

27. The polar of P with respect to the circle x'^-\-y'^ â€” a- touches

the circle {x - of + {y - ^)"â€”b^ ; prove that its locus is the curve given

by the equation {ax + ^y-a^f â€” b^ {x^ + y^) .

28. A tangent is drawn to the circle {x - a)" + y" = b- and a perpen-

dicular tangent to the circle {x + a)'^ + y- = c" ; find the locus of their

point of intersection, and prove that the bisector of the angle between

them always touches one or other of two fixed circles.

XXII.] EXAMPLES. ' 159

29. Ill any circle prove that the perpendicular from any point of

it on the line joining the points of contact of two tangents is a mean

proportional between the perpendiculars from the point upon the two

tangents.

30. From any point on the circle

a;2 + if + 2gx + 2fy + c^0

tangents are drawn to the circle

x^ + ij'^ + 2gx + 2fy + c sin^ a + {g'^+f^) cos^ a = 0;

prove that the angle between them is 2a.

31. The angular points of a triangle are the points

(a cos a, a sin a), (acos/S, asin/3), and (a cos 7, a sin 7);

prove that the coordinates of the orthocentre of the triangle are

a (cos a + cos /3 + cos 7) and a (sin a + sin /3 + sin 7).

Hence prove that if ^i, B, C, and D be four points on a circle the

orthocentres of the four triangles ABC, BCD, CDA, and DAB lie on

a circle.

32. A variable circle passes through the point of intersection

of any two straight lines and cuts ofi from them portions OP and OQ

such that m . OP + n . OQ is equal to unity ; prove that this circle

always passes through a fixed point.

33. Find the length of the common chord of the circles, whose

equations are (x-af + y'^ = a? and x^ + (1/ - h)-=^JP, and prove that the

equation to the circle whose diameter is this common chord is

(a2 + &2) (a;2 + yi-^ ^2ab {bx + aij) .

34. Prove that the length of the common chord of the two circles

whose equations are

(a;-a)2 + (?/-fc)2 = c-2 and {x-hf + iy-af^c^

is V^c- - 2 (a - h)-K

Hence find the condition that the two circles may touch.

35. Find the length of the common chord of the circles

x^ + y^-2ax-4ay -4^0^ = and X' + y'-oax + 4:ay = 0.

Find also the equations of the common tangents and shew that

the length of each is 4a.

36. Find the equations to the common tangents of the circles

(1) x'^ + 7f-2x~&y + 9 = and x- + y-^ + ex-2y + 1 = 0,

(2) x^ + 7/2 = 6-2 and {x - a)^ + y^ = h-.

CHAPTER IX.

SYSTEMS OF CIRCLES.

[This chapter may he omitted hy the student on a first

reading of the subject.]

182. Orthogonal Circles.

said to intersect orthogonally when

the tangents at their points of

intersection are at right angles.

If the two circles intersect at

P, the radii O^P and O.^P., which

are perpendicular to the tangents

at P, must also be at right angles.

Def. Two circles are

Hence

O^Oi=O^P^-vO.JP\

i.e. the square of the distance between the centres must be

equal to the sum of the squares of the radii.

Also the tangent from Oo, to the other circle is equal to

the radius a^,, i.e. if two circles be orthogonal the length of

the tangent drawn from the centre of one circle to the

second circle is equal to the radius of the first.

Either of these two conditions will determine whether

the circles are orthogonal.

The centres of the circles

a;2 + 2/2 + 2f;^ + 2/?/ + c = and a;- + ?/2 + 2^'a; + 2/'y + c' = 0,

are the points {-g, -/) and (-^', -/') ; also the squares of their

radii are g^+f~- c and g'^ +f'^ - c'.

RADICAL AXIS OF TWO CIRCLES. 161

They therefore cut orthogonally if

i.e. if 2gg' + 2ff=c + c'.

1S3. Radical Axis. Def. The radical axis of

two circles is the locus of a point which moves so that the

lengths of the tangents drawn from it to the two circles are

equal.

Let the equations to the circles be

x' + y' + 2gx + 2fi/ + G^0 (1),

and x" + i/^ + 2gjX + 2fy + c^ = (2),

and let (x^, y^ be any point such that the tangents from it

to these circles are equal.

By Art. 168, we have

^1^ + Vx + ^9^1 + 2/2/1 + c = a^i" + 2/r + 2g^x^ + 2fy^ + c^,

i.e. 2x^{g-g,) + 2y,{f-f,) + G-C:, = 0.

But this is the condition that the point (iCj, 2/1) should

lie on the locus

2x{g-g,) + 2y{f-A) + c-c, = (3).

This is therefore the equation to the radical axis, and it

is clearly a straight line.

It is easily seen that the radical axis is perpendicular

to the line joining the centres of the circles. For these

centres are the points (â€” g, -f) and (-^1, â€” /i)- The

"m" of the line ioining them is therefore -^ â€” 7 {,

^ ^ -9x-{-9)

i.e. -tA.

9-9i

The "?7i" of the line (3) is - 7â€” Â§.

The product of these two " m's " is - 1 .

Hence, by Art. 69, the radical axis and the line joining

the centres are perpendicular.

L. 11

162

COORDINATE GEOMETRY.

184. A geometrical construction can be given

for the radical axis of two circles.

R,

^^^

\ >v

~~^^^/

^ ^v

/ \

A.

^ â€”

Ss

/ *

/ \

/^

7\

1

I

' \

o.

o

o.

Fig. 1.

Fig. 2.

If the circles intersect in real points, F and Q^ as in

Fig. 1, the radical axis is clearly the straight line PQ.

For if T be any point on PQ and TR and TS be the

tangents from it to the circles we have, by Euc. iii. 36,

TR'

TS\

TP . TQ

If they do not intersect in real points, as in the second

figure, let their radii be % and a^, and let 2^ be a point such

that the tangents TR and TS are equal in length.

Draw TO perpendicular to 0^0.2-

Since TR^=TS\

we have T0{- - O^R'' = TOi - O^S^

TO'' + 0^0'' - a^- - TO'' + OO.J - ai,

0^0''-00i=^a^-ai,

{0^0 - 00^) (0^0 + 00.) = ai' - a^\

0^0 â€” OOo =""7777^ = ^ constant quantity.

UiU.2

Hence is a fixed point, since it divides the fixed

straight line O^O^ into parts whose difference is constant.

Therefore, since O^OT is a right angle, the locus of T,

i.e. the radical axis, is a fixed straight line perpendicular to

the line joining the centres.

t.e.

%.e.

%.e,

I.e.

RADICAL AXIS. 163

185. If the equations to the circles in Art. 183 be

written in the form aS'=0 and aS" = 0, the equation (3) to

the radical axis may be written S â€” S' = 0, and therefore

the radical axis passes through the common points, real or

imaginary, of the circles S = and aS" = 0.

In the last article we saw that this was true geometri-

cally for the case in which the circles meet in real points.

When the circles do not geometrically intersect, as in

Fig. 2, we must then look upon the straight line TO as

passing through the imaginary points of intersection of the

two circles.

186. The radical axes of tJwee circles^ taken in 2)cdrs^

meet iri a j^oint.

Let the equations to the three circles be

^ = (1),

'S^'^O (2),

and S"=^0 (3).

The radical axis of the circles (1) and (2) is the straight

line

S->S' = (4).

The radical axis of (2) and (3) is the straight line

.S"-.S"'=:0 (5).

If we add equation (5) to equation (4) we shall have the

equation of a straight line through their points of inter-

section.

Hence .S'-aS'"-0 (6)

is a straight line through the intersection of (4) and (5).

But (6) is the radical axis of the circles (3) and (1).

Hence the three radical axes of the three circles, taken

in pairs, meet in a point.

This point is called the Radical Centre of the three

circles.

This may also be easily proved geometrically. For let

the three circles be called A, B, and C, and let the radical

axis of A and B and that of B and C meet in a point 0.

11â€”2

164 COORDINATE GEOMETKY.

By the definition of the radical axis, the tangent from

to the circle A = the tansrent from ^ - -^

to the circle B, and the tangent f V j

from to the circle B = tangent z'^' \ ^^\^^

from it to the circle C. f \ Ip

Hence the tangent from to >v ^/^v^' /^"^^

the circle A = the tangent from it ^^7\ \ \

to the circle C, i.e. is also a I ^^ j-

point on the radical axis of the V J

circles A and C. â€”

187. If S=0 and S' = he the equations of two circles,

the equation of any circle through their 2>oints of inter-

section is S â€” \S'. Also the equation to any circle, such that

the radical axis of it and Sâ€”0 is u = 0, is S + \u ~ 0.

For wherever S =0 and >S" = are both satisfied the

equation S = XS' is clearly satisfied, so that S = \S' is some

locus through the intersections of aS' = and *S"= 0.

Also in both S and S' the coefiicients of x^ and y'^ are

equal and the coefiicient of xy is zero. The same statement

is therefore true for the equation S=XS'. Hence the

proposition.

Again, since u is only of the first degree, therefore in

S + Xu the coefficients of ay^ and y^ are equal and the

coefficient of xy is zero, so that S + Xu = is clearly a circle.

Also it passes through the intersections of *S^ = and u â€” 0.

EXAMPLES. XXIII.

Prove that the following pairs of circles intersect orthogonally :

1. x'^ + y^-2ax + c = and x^ + y^ + 2by -c = 0.

2. x^ + y- - 2ax + 2hy + c = and x^ + y^ + 2hx + 2ay -c = 0.

3. Find the equation to the circle which passes through the origin

and cuts orthogonally each of the circles

x^ + y^-ex + 8 = and x^ + y^-2x-2y = 7.

Find the radical axis of the pairs of circles

4. ic2 + 2/2=144 and x^ + y^-15x + lly = 0.

5. x^ + y^-3x-4:y + 5=0 and Sx^- + By^-7x + 8y + ll=0.

RADICAL AXIS. EXAMPLES. 165

6. x^ + y - xy + Qx-ly + 8 = &nd x^ + ij' -xy-4 = 0,

the axes being inclined at 120Â°.

Find the radical centre of the sets of circles

7. x^+y^ + x + 2y + S = 0, x^ + 2f-h2x + 4:y + 5 = 0,

and x^ + y^-7x-8y-d = 0.

8. (a;-2)2+(z/-3)2 = 36, (re + 3)2 +(7/ + 2)2 = 49,

and (a;-4)2 + (i/ + 5)2=64.

9. Prove that the square of the tangent that can be drawn from

any point on one circle to another circle is equal to twice the product

of the perpendicular distance of the point from the radical axis of the

two circles, and the distance between their centres.

10. Prove that a common tangent to two circles is bisected by the

radical axis.

11. Find the general equation of aU circles any pair of which have

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