If one end of diameter of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ is (7, -5) , then other end of diameter is

( -1, -3)

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( -1, 1)

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( -4, 3)

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(-4, 4)

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Solution

The correct option is B
( -1, 1)

Centre C=(3, -2) one end A=(7,-5) other end = 2C – A = (-1, 1)

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x^{2} + y^{2} - 4 x - 6 y + 11 = 0

Equation of the circle on the latusrectum of $y^{2} = 8 x$ as ends of diameter is

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

x^{2} + y^{2} - 6 x - 8 y + 12 = 0

(4, 6)

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