If circle {x^2} + {y^2} + 6x - 2y + k = 0 bisects circumference of circle {x^2} + {y^2} + 2x - 6y -

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10-1.Circle and System of Circles

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If the circle ${x^2} + {y^2} + 6x - 2y + k = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2x - 6y - 15 = 0,$ then $k =$

A

$21$

B

$-21$

C

$23$

D

$-23$

Solution

(d) $2{g_2}({g_1} – {g_2})\, + 2{f_2}({f_1} – {f_2}) = {c_1} – {c_2}$

$2(1)\,\,(3 – 1) + 2( – 3)\,\,( – 1 + 3) = k + 15$

$4 – 12 = k + 15$ or $ – 8 = k + 15\, $

$\Rightarrow \,k = – 23$.

Standard 11

Mathematics

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