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The locus of the mid points of the chords of the circle $x^{2} + y^{2} + 4 x - 6 y - 12 = 0$ which subtend an angle of $\frac{π}{3}$ at its circumference is

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${(x - 2)}^{2} + {(y + 3)}^{2} = 6.25$

${(x + 2)}^{2} + {(y - 3)}^{2} = 6.25$

${(x + 2)}^{2} + {(y - 3)}^{2} = 18.75$

${(x + 2)}^{2} + {(y + 3)}^{2} = 18.75$

answer is B.

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Detailed Solution

Let $P (h, k)$ be the mid point of chord AB of given circle whose centre $C = (- 2, 3)$

Since $A C P = \frac{π}{3}, 2 P C = A C$

$\Rightarrow 4 ({(h + 2)}^{2} + {(k - 3)}^{2}) = 25$

Locus of $P (h, k)$ is ${(x + 2)}^{2} + {(y - 3)}^{2} = \frac{25}{4} = 6.25$

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