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 the value of $K$ is -

21

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Solution

The correct option is C $- 23$
$s_{1} : x^{2} + y^{2} + 6 x - 2 y + k = 0$

and $s_{2} : x^{2} + y^{2} + 2 x - 6 y - 15 = 0$

then $s_{1} - s_{2} = 0$ given the equation of common chord of $s_{1}$ and $s_{2}$

$4 x + 4 y + k + 15 = 0$

and common chord is a diameter of $x^{2} + y^{2} + 2 x - 6 y - 15 = 0$

then common chord passel through the centre of $s_{2}$

$∴ (- 1, 3)$ in $x + 4 y + k + 15 = 0$

$\Rightarrow - 4 + 12 + k + 15 = 0$

$k = - 23$

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Similar questions

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 _______

x^{2} + y^{2} - 4 x + 6 y + 3 = 0

x^{2} + y^{2} + 6 x - 4 y + k = 0

x^{2} + y^{2} + 6 x + 8 y + a = 0

x^{2} + y^{2} + 2 x - 6 y - b = 0

a + b

x^{2} + y^{2} + 4 x - 6 y + c = 0

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

C_{1} \equiv x^{2} + y^{2} - 6 x - 4 y + 9 = 0

C_{2} \equiv x^{2} + y^{2} - 8 x - 6 y + 23 = 0

C_{1}

C_{2}

C_{1}

C_{2}

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