The shortest distance from $(3, 5)$ to the circle $x^{2} + y^{2} + 2 x - 4 y = 0$ is

2

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3

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2.76

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7

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Solution

The correct option is C $2.76$
$x^{2} + y^{2} + 2 y - 4 y = 0 x^{2} + 2 x + {(1)}^{2} + y^{2} - 4 y + {(2)}^{2} - 1 - 4 = 0$
${(x + 1)}^{2} + {(y - 2)}^{2} = {(\sqrt{5})}^{2} 0 (- 1, 2)$ radius $= \sqrt{5}$
length of where $M (3, 5)$
$\sqrt{4^{2} + 3^{2}} = 5$
So, $O$ is exterior to the circle and the shortest distance is
$5 - \sqrt{5} = 2.76$
(3) option correct

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