The nearest point on the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ from $(- 5, 4)$ is?

(1, 1)

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(- 1, 1)

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(- 1, 2)

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(- 2, 2)

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Solution

The correct option is A $(- 1, 1)$
$\begin{matrix} x^{2} + y^{2} - 6 x + 4 y - 12 = 0 {(x - 3)}^{2} + {(y + 2)}^{2} = 12 + 13 P C = \sqrt{8^{2} + 6^{2}} P C = \sqrt{64 + 36} P C = 10 C R = 5 & P R = 5 (- 5, 4) \end{matrix}$
lies on the circle
PC is normal to circle
R is mid - point of PC
$\begin{matrix} R = (\frac{- 5 + 3}{2}, \frac{4 - 2}{2}) R = (\frac{- 2}{2}, \frac{2}{2}) R = (- 1, 1) \end{matrix}$

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