The other end of the diameter through the point $(- 1, 1)$ on the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ is :

(- 7, 5)

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(- 7, - 5)

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(7, - 5)

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(7, 5)

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Solution

The correct option is C $(7, - 5)$
Given, circle is $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$
Centre of this circle is $(3, - 2)$
One end of the diameter is $(- 1, 1)$
Let the other end of the diameter be $(α, β)$
$∴ \frac{α - 1}{2} = 3, \frac{β + 1}{2} = - 2$
$\Rightarrow α = 7, β = - 5$
$∴$ Other end of the diameter is $(7, - 5)$

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