The number of common tangents for circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 8 x - 6 y - 24 = 0$ is

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Solution

The correct option is A 1
$C_{1} =$ centre of 1st circle $\equiv (0, 0)$ and radius $= r_{1} = 2 u n i t s$
$C_{2} =$ centre of 2nd circle = (4,3) and radius $= r_{2} = \sqrt{16 + 9 + 24} = 7$
$∴ C_{1} C_{2} = \sqrt{(0 - 4)^{2} + (0 - 3)^{2}} = 5$
$| r_{2} - r_{1} | = | 7 - 2 | = 5$
$C_{1} C_{2} = | r_{2} - r_{1} |$
Both circles touch each other internally.

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