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Question

If two circles cut orthogonally, prove that the polar of any point P on the first circle with respect to the second passes through the other end of the diameter of the first circle which goes through P. Hence , (by considering the orthogonal circle of three circles as the locus of a point such that its polars with respect to the circles meet in a point ) prove that the orthogonal circle of three circles, given by the general equation is
∣ ∣x+g1y+f1g1x+f1y+c1x+g2y+f2g2x+f2y+c2x+g3y+f3g3x+f3y+c3∣ ∣=0

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Solution

Let the equation of the first circle be
x2+y2=a2............(1)
and let the other circle be
x2+y2+2gx+2Fy+c=0...........(2)
If these two circles at orthogonality, then
2g×0+2F×0=a2c
c=a2
Hence, equation (2) becomes,
x2+y2+2gx+2Fy+a2=0............(3)

Let, (acosα,asinα be any point on (1). The poles of (acosα,asinα) with respect to (s) will be
(x.acosα+y.asinα+g(x+acosα)+f(y+asinα)+a2=0............(4)
The co-ordinates of the other end of the diameter of the circle (s) of which one end is (acosα,asinα) is clearly.
(acosα,asinα)
Substituting these co-ordinates in the L.H.S of (4)
acosα.acosαasinα.asinα+g(acosα+acosα)+F(asinα+asinα)+a2=0

a2(cos2α+sin2α)+a2=0 =R.H.S

(acosαasinα) satisfies (4) Hence proved.

b) Let the three circle be

x2+y2+2g1x+2F1y+c1=0...........(1)

x2+y2+2g2x+2F2y+c2=0...........(2)

x2+y2+2g3x+2F3y+c3=0...........(3)

Let the co-ordinates of the moving point be (h,k)
The equation of the polar of (h,k) with respect to (1) is

xh+yk+g1(x+h)+F1(y+k)+c1=0
x(h+g1)+y(h+F1)+(g1h+F1k+c1)=0..........(4)

Similarly, the polars or (h,k) with respect to (2) and (3) respectively,
x(h+g2)+y(k+F2)+(g2h+F2k+c2)=0..............(5)

and x(h+g3)+y(k+F3)(g3h+F3k+c3)=0..........(6)

If the circle are orthogonal, then the three polats should pass through one point, so if (4), (5) R (6) are concurrent the required condition will be.

∣ ∣h+g1k+F1g1h+F1k+c1h+g2k+F2g2h+F2k+c2h+g3k+f3g3h+F3k+c3∣ ∣=0

Generalising for (h,k) the required locus is

∣ ∣x+g1y+F1g1x+F1y+c1x+g2y+F2g2x+F2y+c2x+g3y+F3g3x+F3y+c3∣ ∣=0

Hence proved.

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