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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 $x^{2} + y^{2} + 6 x + 6 y + 14 = 0$ and $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ is

A

4

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B

3

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C

2

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D

1

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Solution

The correct option is A $4$
Given circles are
$x^{2} + y^{2} + 6 x + 6 y + 14 = 0$ and $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$
The centre and radius are,
$C_{1} (- 3, - 3), r_{1} = 2 C_{2} (1, 2), r_{2} = 3 C_{1} C_{2} = \sqrt{41} r_{1} + r_{2} = 5 \Rightarrow C_{1} C_{2} > r_{1} + r_{2}$

Hence, the number of common tangents is $4.$

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Number of Common Tangents to Two Circles in Different Conditions

Standard XII Mathematics

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