If one end of a diameter of the circle $x^{2} + y^{2} - 8 x - 14 y + c = 0$ is the point $(- 3, 2)$ , then its other end point is:

(11, 9)

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(12, 11)

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(11, 10)

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(11, 12)

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Solution

The correct option is D $(11, 12)$
$x^{2} + y^{2} - 8 x - 14 y + c = 0$
Centre of circle $= - (\frac{- 8}{2}), - (\frac{- 4}{2}) = (4, 7)$
See the given figure, $\frac{- 3 + x}{2} = 4 - 3 + x = 8 x = 11$ $\frac{2 + y}{2} = 7 2 + y = 14 y = 12$
$B = (x, y) = (11, 12)$

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