If the circle $x^2+y^2+6 x-2 y+k=0$ bisects the circumference of the circle $x^2+y^2+2 x-6 y-15=0$, then $k$…
If the circle $x^2+y^2+6 x-2 y+k=0$ bisects the circumference of the circle $x^2+y^2+2 x-6 y-15=0$, then $k$ is equal to :
21
-21
23
-23
Solution
Given that,
$S_1 \equiv x^2+y^2+6 x-2 y+k=0$
and $\quad S_2 \equiv x^2+y^2+2 x-6 y-15=0$
Since, $S_1$ bisects $S_2$, then
Chord of $S_2=$ Diameter of $S_1$
Equation of the chord is $S_1-S_2=0$
$\left(x^2+y^2+6 x-2 y+k\right)$
$-\left(x^2+y^2+2 x-6 y-15\right)=0$
$\Rightarrow \quad 4 x+4 y+k+15=0$
Centre of the circle of $S_2=(-1,3)$
Since, equation of the chord passes through $(-1,3)$, then
$4(-1)+4(3)+k+15=0$
$\Rightarrow \quad-4+12+k+15=0$
$\Rightarrow \quad k=-23$