Question

Suppose $S_1$ and $S_2$ are two unequal circles; $AB$ and $CD$ are the direct common tangents to these circles. A transverse common tangent $PQ$ cuts $AB$ in $R$ and $CD$ in $S$. If $AB=10$, then $RS$ is:

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

Answer: (C)

$10$

We know that the lengths of two tangents from a point to the circle are equal.

Let $RQ=RB=x$ and $SC=SP=y$ as shown in the figure.

Given that $AB=10$, we have $AR=RP=10-x$.

As $RP=10-x$ and $RQ=x$, we get, $PQ=10-2x$.

$SQ=SD=10-2x+y$

Also, as the lengths of direct common tangents are equal, $AB=CD=10$.

$CD=CS+SD\Rightarrow 10=y+10-2x+y\Rightarrow x=y$

$RS=RQ+PQ+PS=x+10-2x+y=10-x+y=10$