Q65 / 160 · AP EAMCET 2003 · Mathematics

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 :

  • A
    21
  • B
    -21
  • C
    23
  • Correct answer
    D
    -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 $4(-1)+4(3)+k+15=0$ $\Rightarrow \quad-4+12+k+15=0$ $\Rightarrow \quad k=-23$

Hence, the correct answer is option: 'D'.

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