Circle problem in which i wanted to find value of $PQ$

Question

Find a value of $PQ$:

Let the radius of bigger circle be $R$, and that of the smaller circles be $r_1$ and $r_2$.

Hence we can find value of $R = 10$. I don't know how to proceed further to find the Value of $r_1$ and $r_2$.

Answer 1

Let $AP =x$, $PQ = y$ and $QB = z$, so by Pythagoras theorem we have $$x+y+z = 20$$

Since $AC = AQ$ and $BP = BC$ we have also: $$x+y=12$$ $$y+z=16$$

so $PQ = y = 8$.

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I used a following lemmas:

Lemma 1: $A,H,G$ are collinear.

Proof: Observe a homothety at $G$ wich takes smaller circle to a bigger and let $H'$ be a image of $H$. Then $H'$ is on bigger circle and tangent $CH$ goes to parallel tangent on bigger circle, so it can go only through $A$, so $H'=A$ and thus $A,H,G$ are collinear.

Lemma 2: $AF = AC$

Proof: Suppose $A,H,G$ are collinear. Since $BGHD$ is cyclic we can use the power of the point $A$: $$ AH \cdot AG = AD\cdot AB$$ If we use the power of the point $A$ with respect to smaller circle we have:

$$ AH\cdot AG = AF^2$$

And if we use the power of the point $A$ with respect to circle $(BCD)$ which is tangent to $AC$ we have: $$AD\cdot AB = AC^2$$

So $AC = AF$.

Answer 2

I am using the lemmas provided by @greedoid . Then the calculation of PQ can be simplified as:-

$PQ = AQ + BP - AB = 12 + 16 - 20 = 8$

Answer 3

Using the power of the point $O$ with respect to circle $O_1$ we have

\begin{align} |OT_2|\cdot|OT_1|&=|OP|^2 ,\\ \text{or }\quad R(R-2r_1)&=(R-(|AD|-r_1))^2 ,\\ r_1&= \sqrt{2R(2R-|AD|)}-(2R-|AD|) \tag{1}\label{1} . \end{align}

Similarly,

\begin{align} r_2&=\sqrt{2R\cdot|AD|}-|AD| \tag{2}\label{2} . \end{align}

Note that given $a=16=4\cdot4$, $b=12=4\cdot3$ means that $\triangle ABC$ is a scaled version of the famous $3-4-5$ triangle, so $R$ and $|AD|$ can be easily calculated,

\begin{align} R&=4\cdot\frac 52=10 ,\\ |AD|&=4\cdot3\cdot\frac 35=\frac{36}5 ,\\ r_1&=\frac{16}5 ,\\ r_2&=\frac{24}5 ,\\ |PQ|&=8 . \end{align}