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The greatest ...

Question

The greatest distance of the point P $(10, 7)$ to the circle $x^{2} + y^{2} - 4 x - 2 y - 20 = 0$ is:

10

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15

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5

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12

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Solution

The correct option is B $15$
$x^{2} + y^{2} - 4 x - 2 y - 20 = 0$
Centre $C \equiv (2, 1),$ Radius $R = 5$
Greatest distance of $P = P C + R = \sqrt{8^{2} + 6^{2}} + 5$
$= 10 + 5 = 15$

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