The shortest distance from the point $(2, - 7)$ to the circle $x^{2} + y^{2} - 14 x - 10 y - 151 = 0$ is

1

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2

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3

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4

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Solution

The correct option is D $3$
$P (- 7, 2)$

$x^{2} + y^{2} - 10 x - 14 y - 151 = 0$

$C (- g, - f) = C (5, 7)$

$c = - 151$

$r = \sqrt{5^{2} + 7^{2} - (- 151)} = 15$

$P (- 7, 2), C (5, 7)$

$C P = \sqrt{(5 + 7)^{2} + (7 - 2)^{2}} = 13$

The shortest distance

$= r - C P$

$= 15 - 13 = 2$

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