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

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.