Question

If the tangents are drawn to the circle $x^2+y^2=12$ at the point where it meets the circle $x^2+y^2-5x+3y-2=0$, then the point of intersection of these tangents is

• A. $(6,-\frac{18}{5})$
• B. $(-6,\frac{18}{5})$
• C. $(7,-\frac{18}{6})$
• D. $(8,-\frac{18}{5})$

Answer 1

The correct option is A $(6,-\frac{18}{5})$


Let $(h,k)$ be the point of intersection of the tangents.
Then the chord of contact of tangents is the common chord of the circle $x^2+y^2=12$ and $x^2+y^2-5x+3y-2=0.$
i.e., $S_1-S_2=0$
$\Rightarrow 5x-3y-10=0$

Also, the equation of the chord of contact w.r.t. $P$ is $T=0$
i.e., $hx+ky-12=0$
Equations $hx+ky-12=0$ and $5x-3y-10=0$ represent the same line.
$\therefore \frac{h}{5}=\frac{k}{-3}=\frac{-12}{-10}$
$\Rightarrow h=6,k=\frac{-18}{5}$
Hence, the required point is $(6,-\frac{18}{5})$.