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!

For the two circles to touch at exactly one point, the distance between the centers of the two circles should be the sum of the radii.
$$4r^2=(0-x_0)^2+(L-mx_0)^2.$$
Now solve for $m$.
