Question

 $\frac{\mathrm{C}}{\mathrm{C}-3}=\frac{2}{1}$

2C – 6 = C

 $\Rightarrow$ C = 6

 Hence, A (6, 0)

Equation of a line through A is y =m (x – 6)         .......(3)

 $\Rightarrow$ mx – y – 6m = 0

equation (3) is tangent to both, hence   $\left|\frac{3\mathrm{m}-6\mathrm{m}}{\sqrt{1+{\mathrm{m}}^2}}\right|$  = 1

 $\Rightarrow$ 9m² = 1 + m²  $\Rightarrow \mathrm{m}=\frac{1}{2\sqrt{2}}\mathrm{or}\frac{-1}{2\sqrt{2}}$

 $\therefore$ Common tangent is y =  $\frac{1}{2\sqrt{2}}$  (x – 6)  $\Rightarrow \mathrm{x}-\sqrt{2}$ 2y = 6             

or – 2$\sqrt{2}$ y = x – 6 Þ x + 2$\sqrt{2}$  = 6.

Equation of tangent at P($\sqrt{3}$ , 1)

on x² + y² = 4 is  $\sqrt{3}$ x + y = 4                              .......(1)             

line perpendicular to equation (1)

 $\mathrm{x}-\sqrt{3}\mathrm{y}+\mathrm{\lambda }=0$                                                ........(2)

Perpendicular from (3, 0) on (2) = unity

 $\left|\frac{3+\mathrm{\lambda }}{2}\right|=1\Rightarrow \mathrm{\lambda }+3=\pm 2$

  $\Rightarrow \mathrm{\lambda }$ = – 1 or $\mathrm{\lambda }$ = 5

 $\therefore$ Equation of L is x –  $\sqrt{3}$ y – 1 = 0