I am doing an exercise on system of circles:
Find the equations of the circles which are orthogonal to all the circles of the system $x^2+y^2+2ax+2by-2\lambda(ax-by)=0\tag*{}$ where $\lambda$ is a varying parameter and $a$, $b$ are constants.
I recognized that $x^2+y^2+2ax+2by-2\lambda(ax-by)=0\tag*{}$ is a system of coaxial circles with the common radical axis $ax-by=0$.
As a consequence of the above, the center of a circle (call it $S$) which is orthogonal to the system will lie on $ax-by=0$. I concluded this from a theorem in my previous exercise that:
If two circles cut a third circle orthogonally then their radical axis/common chord/common tangent will pass through the center of the third circle).
If I let the equation of $S$ be $x^2+y^2+2gx+2fy+c=0$, by condition of orthogonality, we have: $2a(1-\lambda)g+2b(1+\lambda)f=c\tag*{}$ for all $\lambda \in \mathbb R$.
If I set $\lambda=1$, I get $c=4bf$ and if $\lambda=-1$, I get $c=4ag$. This means that $bf=ag$ (which is the same conclusion one would get using the fact that the center of $S$ lies on $ax-by=0$ so it's not useful.)
My goal remains to eliminate $f$, $g$, $c$ and also find all circles which are orthogonal to the system. Am I on the right track? I am not sure how to proceed.
Update:
I realized that there is no need to eliminate $f$, $g$ and $c$ at the same time because the orthogonal circles themselves will form another system of coaxial circles with some varying parameter in the linear part of the equation.
I chose $g$ to be a parameter then put $f=ag/b$ and $c=4ag$, the equation of $S$ becomes: $x^2+y^2+2gx+2\left(\frac{ag}{b}\right)y+4ag=0\tag*{}$
Replace $g$ by $b\lambda$, $$ \boxed{x^2+y^2+2\lambda(bx+ay+2ab)=0}$$
I think that's the correct answer, here's a visual demonstration. I am not sure if all the orthogonal circles are covered in this equation.