Question

Tangents are drawn from a point $P$ on the circle $C:x^2+y^2=a^2$ to the circle $C_1:x^2+y^2=b^2$. These tangents cut the circle $C$ at $Q$ and $R$. If $QR$ touches $C_1$, then the area of $△PQR$ is :

• A. $2\sqrt{3}ab$
• B. $\frac{3\sqrt{3}}{2}ab$
• C. $\frac{3\sqrt{3}}{4}{a}^2$
• D. $3\sqrt{3}{b}^2$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\frac{3\sqrt{3}}{2}ab$
C $\frac{3\sqrt{3}}{4}{a}^2$
D $3\sqrt{3}{b}^2$

$PD=PE,RF=RE,QD=QF......$ ($\because$ tangents drawn to the same circle)
Also, $PE=ER$ ($\because$ $OPR$ is an isosceles triangle).
Hence, we get $PQ=QR=PR$. So $PQR$ is an equilateral triangle.
$\Rightarrow a=2b$
$\Rightarrow$ height of the triangle is $3b$.
Hence, the length of the side the triangle is $\frac{2}{\sqrt{3}}3b=2\sqrt{3}b$
$\Rightarrow$ the area of the triangle is $\frac{\sqrt{3}}{4}(2\sqrt{3}b{)}^2=3\sqrt{3}{b}^2$
Using $a=2b$, we get the area of the triangle is also equal to $\frac{3\sqrt{3}}{4}{a}^2$
Similarly, in terms of $a$ and $b$, area$=\frac{3\sqrt{3}}{2}ab$
Hence, options 'B', 'C' and 'D' are correct.