Tangents are drawn from a point P on the circle $C:x^2+y^2=a^2$  to the circle $C_1:x^2+y^2=b^2$ 
if these tangents cut the circle C at Q and R and if QR is a tangent to the circle $C_1$  . Then
The area of triangle PQR is

Conceptual

$$\begin{gathered} \mathrm{PD}=\mathrm{PE},\mathrm{RF}=\mathrm{RE},\mathrm{QD}=\mathrm{QF}\dots \dots (\because \text{ tangents drawn to the same circle }) \\ \text{ Also, }\mathrm{PE}=\mathrm{ER}(\because \mathrm{OPR}\text{ is an isosceles triangle). } \end{gathered}$$

$$\begin{gathered} PQ=QR=PR \\ \text{ So }\mathrm{PQR}\text{ is an equilateral triangle. } \end{gathered}$$

$$\begin{gathered} \Rightarrow a=2b \\ \Rightarrow \text{ height of the triangle is }3b\text{. } \\ \text{ Hence, the length of the side the triangle is }\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}}3 \mathrm{b}=2\sqrt{3} \mathrm{b} \\ \Rightarrow \text{ the area of the triangle is }\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 4}}(2\sqrt{3}\mathrm{b}{)}^2=3\sqrt{3}{\mathrm{b}}^2 \\ \text{ Using }\mathrm{a}=2\mathrm{b}\text{, we get the area of the triangle is also equal to }\frac{{\displaystyle 3\sqrt{3}}}{{\displaystyle 4}}{\mathrm{a}}^2 \\ \text{ Similarly, in terms of }\mathrm{a}\text{ and }\mathrm{b}\text{, area }=\frac{{\displaystyle 3\sqrt{3}}}{{\displaystyle 2}}\mathrm{ab} \end{gathered}$$