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Number of Common Tangents to Two Circles in Different Conditions

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Question

The number of common tangents to the circles $x^{2} + y^{2} - 6 x - 14 y + 48 = 0$ and $x^{2} + y^{2} - 6 x = 0$ is

A

1

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B

2

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C

0

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D

4

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Solution

The correct option is D $4$
$C_{1} = (3, 7), r_{1} = \sqrt{9 + 49 - 48} = \sqrt{10}, C_{2} = (3, 0), r_{2} = \sqrt{3^{2}} = 3$
$C_{1} C_{2} = 7, r_{1} + r_{2} = 3 + \sqrt{10} < 3 + 4 = 7$
Clearly $C_{1} C_{2} > r_{1} + r_{2}$ .
Hence there will be four common tangents.

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