Power of a point tangents

Question

Two circles $A$ and $B$ with centers $P$ and $Q$, respectively, are externally tangent to each other. The power of point $P$ with respect to circle $B$ is $8$. The power of point $Q$ with respect to circle $A$ is $15$. What is the ratio of the areas of circle $A$ and circle $B$?

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How do I go about solving this problem? Is there perhaps a formula for ratios of areas of circles given different powers of points?

Answer 1

$PT^2=8,QU^2=15$ for the secant-tangent theorem $$ \begin{cases} r(r+2R)=8\\ R(R+2r)=15\\ \end{cases} $$

$$ \begin{cases} r^2+2rR=8\to -15r^2-30rR=-120\\ R^2+2rR=15\to8R^2+16rR=120\\ \end{cases} $$ Add the two equations $$8R^2-14rR-15r^2=0$$ Divide all terms by $r^2$ and set $r/R=z$ $$8-14z-15z^2=0\to z=\frac{2}{5};\;\left(z= -\frac{3}{4}\right)$$

$$\frac{r}{R}=\frac{2}{5}$$ $$\frac{Area_A}{Area_B}=\frac{4}{25}$$

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Answer 2

Power of a point $X$ wrt circle $(O,r)$ is $OX^2-r^2$

Given is $$(a+b)^2-b^2=8 \quad , \quad (a+b)^2-a^2=15$$

Define $x:= a/b$ $$\dfrac{(a+b)^2-a^2}{(a+b)^2-b^2}=\dfrac{15}{8}$$

$$\Rightarrow \dfrac{(x+1)^2-x^2}{(x+1)^2-1^2}=\dfrac{15}{8}$$

$$\Rightarrow x=\dfrac{2}{5}$$