Circles $C_1,C_2$ have radii $3$ and $7$ respectively. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

Question

Circles $C_1,C_2$ have radii $3$ and $7$ respectively. The circles intersect at distinct $A$ and $B$. A point $P$ outside $C_2$ lies on the line determined by $A$ and $B$ at a distance $5$ from center $C_1$. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

What I Tried: Here is a picture in Geogebra :-

I have no good idea on how to start this. I cannot, for example, find the lengths of $C_1C$ and $CC_2$ from pure deduction, ($C_1$ and $C_2$ are centres of the circles) and even if Pythagorean Theorem may look like they can work, I cannot understand how, because I don't have many required lengths. I also tried similarity but in fact I could not find any similar triangles here.

Can anyone help me? Thank You.

Answer 1

$$ |PQ|^2=|PA||PB|= (|PC_1|+r_1)(|PC_1|-r_1) $$

Answer 2

There has be a better way, but here is a solution by bashing Pythagoreas' Theorem. By extracting the information of lengths of radii from the circles, we end up with the following figure (here $c=7$, but the length of $PQ$ is independent of $c$):

By Pythagoreas' Theorem:

$$C_1C = \sqrt{9-a^2}, \quad CC_2 = \sqrt{c^2-a^2}, \quad CP = \sqrt{25 - (9-a^2)} = a+b$$ $$(a+b)^2 = 16+a^2$$ $$PQ = \sqrt{PC_2^2-c^2}=\sqrt{((a+b)^2+(c^2-a^2))-c^2}=\sqrt{(a+b)^2-a^2} = \sqrt{16}=4$$