Question

The line $3 x-y+k=0$ touches the circle $x^{2}+y^{2}+4 x-6 y+3=0 .$ If $k_{1}, k_{2}\left(k_{1}< k_{2}\right)$ are the two values of $k$, then the equation of the chord of contact of the point $\left(k_{1}, k_{2}\right)$ with respect to the given circle is

• A. $19 x+y-18=0$
• B. $x+19 y-3=0$
• C. $x+16 y-56=0$
• D. $20 x+18 y-7=0$

Answer 1

Answer: (C)

Line $3 x-y+k=0$ touches the circle
$x^{2}+y^{2}+4 x-6 y+3=0$ having
Centre $(-2,3), r=\sqrt{4+9-3}=\sqrt{10}$
$\because r=\left|\frac{-6-3+k}{\sqrt{10}}\right|$

$10=|-9+k| $
$\Rightarrow \pm 10=k-9$
$\therefore k =-1,19 $
$k_{1} =-1, k_{2}=19$
Equation of chord of contact of $(-1,19)$ to the circle
$-x+19 y+2(x-1)-3(y+19)+3=0 $
$\Rightarrow x+16 y-56=0$