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=a2−c
⇒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.acosα−asinα.asinα+g(−acosα+acosα)+F(−asinα+asinα)+a2=0
⇒−a2(cos2α+sin2α)+a2=0 =R.H.S
∴(−acosα−asinα) 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.