There is something misunderstanding with that question that I think it to have inadequte context or information (obviously for my little knowledge). So I couldn't solve the problem.
SOURCE: BANGLADESH MATH OLYMPIAD
$PE$ is a tangent of the below diagram. The diameter of small circle is equal to the radius of the large circle and $BC$ is the diameter of the small circle. If $PA$ = $12$, $\frac{AB}{CD}$ = $\frac{1}{2}$, then what is the value of $PC$ ?
I denoted the center of small circle $H$ and drew three altitude lines $AG$, $HE$ and $DI$ from the vertices $A$ , $H$ and $DI$ and also denoted $AB$ = $x$ and $BH$ = $y$.
So, $HE$ = $y$. $\triangle AGP$ $\sim$ $\triangle HEP$ $\sim$ $\triangle DIP$. So, I showed the relation of $AG$ and $DI$ with $HI$ with the help of their proportion of length because of their similarity. But I was unable to show relation with the rest property and I couldn't find the value of $x$ and $y$.
I think I went into messed situation for solving the above problem. I don't think so even my used method is going to the right direction.
It will be very helpful for me if someone please tells me in which way should I go or where is my mistake to find the right way. Thanks in advance.
Let $AB = x$ and $CD = 2x$
You can use the power of the point with respect to both circle and point $P$:
$$ PA\cdot PD (=PE^2)= PB\cdot PC$$ $$\implies 12(12+x+2r+2x)= (12+x)(12+x+2r)$$
so $$x+2r = 12\implies PC = 12+x+2r = 24$$
Note: Information about relation betwen both radius of the circles is irrelevant.