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The number of common tangents that can be drawn to the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is

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1 $= 4 .$

answer is A.

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Detailed Solution

For the circle $x^{2} + y^{2} = 1;$ Centre $= C_{1} (0, 0), radius r_{1} = 1 .$

For the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0;$ Centre $C_{2} = (1, 3),$ radius $r_{2} = \sqrt{1 + 9 - 6} = 2 .$

$C_{1} C_{2} = \sqrt{1 + 9} = \sqrt{10} > r_{1} + r_{2} . ∴ o n e c i r c l e c o m p l e t e l y l i e s o u t s i d e$

The number of common tangents $= 4 .$

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