Finding product of the radii of two circles

Question

If two circles touch the line $y=x$ and $y=mx$ and touch each other at the point $P(3,6)$, than find product of their radii.

I tried a lot but always got stucked. Thank you in advance.

Answer 1

In the following, all geometric facts will be stated without proofs.

Let $L_1 : y = x$, and $L_3 : y = mx$.

For the two circles to observe the given, $L_2$, the line of centers, must be the internal angle bisector of $L_1$ and $L_3$. Then, $L_2$ is of the form $y = Nx$ for some N. Since P(3, 6) must also ly on $L_2$, therefore, $N = 2$. That is $L_2 : y = 2x$.

Let Q be the midpoint of HK. The red dotted circle using Q as center with radius $= \dfrac {R+r}{2}$ will touch $L_1$ at X. X can be found by:-

(1) Use point-slope form to setup the equation of PX (which is normal to $L_2$); and (2) Solve PX and $L_1$ to get X = (5, 5).

Hence, $(PX)^2 = 5$.

By power of a point, $Rr = (PX)^2$