Question

If $C_1:x^2+y^2=(3+2\sqrt{2}{)}^2$ be a circle and PA and PB are pair of tangents on $C_1$, where P is any point on the director circle of $C_1$, then the radius of smallest circle which touch $C_1$ externally and also the two tangents $PA$ and $PB$ is

• A. $2\sqrt{3}-3$
• B. $2\sqrt{2}-2$
• C. $2\sqrt{2}-1$
• D. $1$

Answer 1

The correct option is D $1$

$AQ=3+2\sqrt{2}$ $(\because given)$
$PQ=3\sqrt{2}+4$ $(\because \angle QPA=\frac{\pi }{4})$
Let r be required radius
$\therefore 3\sqrt{2}+4=3+2\sqrt{2}+r+r\sqrt{2}$ $(\because \angle RPD=\frac{\pi }{4})$
$PQ$ $=$radius of bigger circle $+r+RP$
$\sqrt{2}+1=r(1+\sqrt{2})\Rightarrow r=1$