Find the radius of a circle touching two circle $x^2+y^2+3\sqrt{2}(x+y)=0$ and $x^2+y^2+5\sqrt{2}(x+y)=0$ and also touching the common diameter of the two given circles.
The two circles touch internally and the common diameter is $x-y=0$.
Let centre of required circle be $(h,k)$
$$\sqrt{\left(h-\frac{3}{\sqrt{2}}\right)^2+\left(k-\frac{3}{\sqrt{2}}\right)^2}=r+3$$
$$\sqrt{\left(h-\frac{5}{\sqrt{2}}\right)^2+\left(k-\frac{5}{\sqrt{2}}\right)^2}=5-r$$
and
$$r=\frac{h-k}{\sqrt{2}}$$
assuming $h>k$
Squaring,
$$h^2+k^2+3\sqrt{2}h+3\sqrt{2}k=r^2+6r$$ $$h^2+k^2+5\sqrt{2}h+5\sqrt{2}k=r^2-10r$$ substituting the value of $r$ and subtracting the two equations, $$k=\frac{5h}{3}$$
But $h>k$. Where am I making the mistake?
I think there are two such circles which are mirror immages about $y=x$. One of them has $h>k$.
I am posting a diagram based on my understanding of the question.