From a point $P$ on the circle $x^2+y^2-4 x-6 y+9=0$, a pair of tangents $P Q$ and $P R$ are drawn touching the circle $x^2+y^2-4 x-6 y+12=0$ at $Q$ and $R$. If $C$ is the centre of the concentric circles, then the area of the $\triangle C Q R$ (in sq. units) is

• A

  $\frac{1}{2}$
• B

  $\frac{\sqrt{3}}{2}$
• Correct answer
  C

  $\frac{\sqrt{3}}{4}$
• D

  $\frac{3}{4}$

Solution

Given equation of circle, $\begin{array}{r} x^2+y^2-4 x-6 y+9=0 \\ R=P C=\sqrt{4+9-9}=2 \\ x^2+y^2-4 x-6 y+12=0 \\ r=Q C=\sqrt{4+9-12}=1 \end{array}$
In $\triangle P Q C$, $\cos \theta=\frac{Q C}{P C}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$ $\therefore$ Area of $\triangle Q R C$, $=\frac{1}{2}(C Q)^2 \times \sin 2 \theta=\frac{1}{2} \times 1 \times \sin \frac{2 \pi}{3}=\frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$

Hence, the correct answer is option: 'C'.