The maximum distance of the point $(4, 4)$ from the circle $x^{2} + y^{2} - 2 x - 15 = 0$ is

10

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9

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5

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none of these

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Solution

The correct option is B $9$
Max distance $= C P + r$

Where C is the center, P is the point , $r - r a d i u s$

$C = (1, 0)$ , $r = \sqrt{1 + 15} = 4$

$P = (4, 4)$

$d = \sqrt{3^{2} + 4^{2}} + 4 = 5 + 4 = 9$

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