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Number of Common Tangents to Two Circles in Different Conditions

The number of...

Question

The number of common tangents to the following pairs of circles $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ and $x^{2} + y^{2} + 4 x - 2 y + 4 = 0$ is

A

0

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B

1

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C

2

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D

3

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Solution

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

Therefore, one circle lies completely inside the other.
Hence, the number of common tangents is $0.$

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x^{2} + y^{2} + 4 x - 2 y + 4 = 0

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Number of Common Tangents to Two Circles in Different Conditions

Standard XIII Mathematics

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