The circles $x^{2} + y^{2} - 2 x - 4 y - 20 = 0, x^{2} + y^{2} + 4 x - 2 y + 4$ are

one lies out side the other

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one lies completely inside the other

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touch eternally

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touch intersally

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Solution

The correct option is A one lies out side the other
$S_{1} = x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ .
$c_{1} = (2, 4); r_{1} = \sqrt{{(- 2)}^{2} + {(- 4)}^{2} + 20} = 6.32$ unit
$S_{2} = x^{2} + y^{2} + 4 x - 2 y + 4 = 0$ .
$c_{2} = (- 4, 2); r_{2} = \sqrt{4^{2} + {(- 2)}^{2} - 4} = 4$ unit
Now, $r_{1} + r_{2} = 10.32$ unit.
Distance between $c_{1}$ and $c_{2} = \sqrt{{(2 + 4)}^{2} + {(4 - 2)}^{2}} = 6.32$ unit
Hence , we find $¯ c_{1} c_{2} < | r_{1} + r_{2} |$
i.e smaller circle lies inside another big circle. So, one lies outside the others.

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