Question

Prove that as $\lambda$ varies the circles $x^2+y^2+2ax+2by+2\lambda (ax-by)=0$
for a coaxial system. Find the equation of radical axis and also the equation of the circle which are orthogonal to the circles of the above system.

Answer 1

$S+\lambda P=0$ forms a coaxial system, the radical axis being $P=0$ for all values of $\lambda$. Let $g,f,c$ be the circle orthogonal to $S+\lambda P=0$. Applying the condition of orthogonality, we get
$2a(1+\lambda )g+2b(1-\lambda )f=0+c$
or $2(ag-bf)\lambda +2(ag+bf)-c=0$
Above relation holds for all values of $\lambda$ and hence
$2(ag-bf)=0,2(ag+bf)-c=0$
$\therefore g=c/4a,f=c/4b$.
$x^2+y^2+\left(c/2a\right)x+\left(c/2b\right)y+c=0$.