For the two circles $x^{2} + y^{2} = 16$ and $x^{2} + y^{2} - 2 y = 0$ there is/are

One pair of common tangents

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Only one common tangent

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Three common tangents

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No common tangent

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Solution

The correct option is C No common tangent
The centres and radii of given circles are $C_{1} (0, 0), r_{1} = 4$ and $C_{2} (0, 1), r_{2} = \sqrt{0 + 1} = 1$
Now, $C_{1} C_{2} = \sqrt{0 + {(0 - 1)}^{2}} = 1$
and $r_{1} - r_{2} = 4 - 1 = 3$
$∴ C_{1} C_{2} < r_{1} - r_{2}$
Hence, second circle lies inside the first circle.

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x^{2} + y^{2} - 4 x - 2 y + 4 = 0

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C_{1} \equiv x^{2} + y^{2} - 4 = 0

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