A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^{2}+y^{2}-1$. Question: A common tangent of the two circles is

• A

  $x=4$
• B

  $y=2$
• C

  $x+\sqrt{3} y=4$
• Correct answer
  D

  $x+2 \sqrt{2} y=6$

Solution

From the figure it is clear that the intersection point of two direct common tangents lies on $x$-axis. Also $\Delta P T_{1} C_{1} \sim \Delta P T_{2} C_{2}$ $\therefore \quad P C_{1}: P C_{2}=2: 1$ or $P$ divides $C_{1} C_{2}$ in the ratio $2: 1$ externally $\therefore \quad$ Coordinates of $P$ are $(6,0)$. Let the equation of tangent through $P$ be $y=m(x-6)$ Asit touches $x^{2}+y^{2}=4$ $\therefore\left|\frac{6 m}{\sqrt{m^{2}+1}}\right|=2 \Rightarrow 36 m^{2}=4\left(m^{2}+1\right)$ $\therefore \quad m=\pm \frac{1}{2 \sqrt{2}}$ $\therefore \quad$ Equations of common tangents are $y=\pm \frac{1}{2 \sqrt{2}}(x-6)$ Also $x=2$ is the common tangent to the two circles.

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