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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 circles $x^{2} + y^{2} - 4 x + 2 y - 4 = 0$ and $x^{2} + y^{2} + 2 x - 6 y + 6 = 0$ is

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is C $3$
Given circles are
$x^{2} + y^{2} - 4 x + 2 y - 4 = 0$ and $x^{2} + y^{2} + 2 x - 6 y + 6 = 0$
The centre and radius are,
$C_{1} = (2, - 1), r_{1} = 3 C_{2} = (- 1, 3), r_{2} = 2 C_{1} C_{2} = \sqrt{9 + 16} = 5 \Rightarrow C_{1} C_{2} = r_{1} + r_{2}$

Therefore, the given circles touch each other externally.
Hence, the number of common tangents is $3.$

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Number of Common Tangents to Two Circles in Different Conditions

Standard XII Mathematics

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