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Q. Two circles are
$S_1 \equiv(x+3)^2+y^2=9 $
$ S_2 \equiv(x-5)^2+y^2=16$
with centres $C_1 \& C_2$
A direct common tangent is drawn from a point $P$ on $x$-axis which touches $S_1$ & $S_2$ at $Q \& R$, respectively. Find the ratio of area of $\triangle PQC _1 \& \triangle PRC _2$.

Conic Sections

Solution:

$\triangle PQC_1$, and $\triangle PRC _2$ are similar
image
$\therefore \frac{\text { Area of } \Delta PQC _1}{\text { Area of } \Delta PRC _2}=\frac{\frac{1}{2} \times PQ \times r _1}{\frac{1}{2} \times PR \times r _2}$
$=\frac{ PQ }{ PR } \cdot \frac{r_1}{r_2}=\frac{r_1}{r_2} \cdot \frac{r_1}{r_2}=\frac{r_1^2}{r_2^2}=\frac{9}{16}$