Finding the centre of a circle under a specific condition Question: Consider a circle, say $\mathscr{C}_1$ with the equation $x^2 + (y-L)^2=r^2$. A second circle, say $\mathscr{C}_2,$ with equal radii that has a centre $(x_0,y_0)$ which lies on the line $y=mx$. Find an expression for $x_0$ and $y_0$, in terms of $L$, $r$ and $m$, such that $\mathscr{C}_1$ and $\mathscr{C}_2$ touch at one point.
My Attempts:

I had attempted to find an expression that would allow for the discriminant to be zero in order for the two circles to only touch once. I ended up with $m = \dfrac{1}{2} (2L - 2\sqrt{10r-x_0}$) although this does not seem to be correct. I arrived at this answer by solving $x^2 + (y-L)^2 = r^2$ and $(x-x_0)^2 + (y-y_0)^2=r^2$ although I am nearly certain that I have made a mistake.
I have also considered using approximate to see if I can identify a relation however as of right now, I have been entirely unsuccessful.

Any help or guidance would be greatly apprecaited!

Hint: Distance between the centers is equal to sum or difference of radii of both circles depending upon whether it touches internally or externally. See this.
