Question

The two circles $x^2+y^2+2ax-c^2=0$ and $x^2+y^2+2abx-c^2=0$ meet in points P and Q. Parallel lines are drawn through P and Q meet the circles in point R and S. Prove that the locus of mid-point of RS is the circle.
$x^2+y^2+(a+b)x=0$

Answer 1

The two circles meet at P (0,c), Q (0,-c). Any time through P is
y = mx + c .........(1)
where m is variable and c is intercept on y-axis and parallel line through Q is
y = mx - c
(1) meets S_1 in point R
$Ris[\frac{-2(a+mc)}{1+m^2},\frac{-2m(a+mc)}{1+m^2}+c]$
(2) meets $S_2$ in point $S,c\to -c,m$ remains same ,$a\to b$
$\therefore Sis[\frac{-2(b-mc)}{1+m^2},\frac{-2m(b-mc)}{1+m^2}-c]$
If (x,y) be the mid-point of RS, then
$2x=\frac{-2(a+b)}{1+m^2},2y=\frac{-m(a+b)}{1+m^2}$
In order to find the locus we have to eliminate the variable m.Dividing we get m = y/x.Putting in any, we get
$2x(1+\frac{y^2}{x^2})=-2(a+b)$
or $x^2+y^2+(a+b)x=0$