Question

Tangents are drawn to the circle $x^2+y^2=12$ at the points where it is met by the circle $x^2+y^2-5x+3y-2=0$; find the point of intersection of these tangents.

Answer 1

$x^2+y^2=\mathrm{12..........}\left(1\right)$

$x^2+y^2-5x+3y-2=\mathrm{0.........}\left(2\right)$

Let both the tangents intersect at P.

AB is chord of both the circles

so $x{x}_1+y{y}_1-12=\mathrm{0.......}\left(3\right)$

eq(1)-eq(2)=0

$\begin{array}{c}x{x}_1+y{y}_1-12=\mathrm{0.......}\left(3\right)\\ x^2+y^2-12-x^2-y^2+5x-3y+2=0\\ 5x-3y-10=\mathrm{0........}\left(4\right)\end{array}$

(3) and (4) represent same line

$\begin{array}{c}\frac{x_1}{5}=\frac{y_1}{-3}=\frac{-12}{-10}=\frac{6}{5}\\ x_1=6,y_1=\frac{-18}{5}\end{array}$