The number of common tangents that can be drawn to the circles $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ and $x^{2} + y^{2} = 2 x + 2 y + 1 = 0$ is

1

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2

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3

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4

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Solution

The correct option is C $3$
$S_{1} = (x - 2)^{2} + (y - 3)^{2} = 16$

$S_{2} = (x + 1)^{2} + (y + 1)^{2} = 1$

$⟹ C_{1} = (2, 3), r_{1} = \sqrt{1} 6 = 4$

$C_{2} = (- 1, - 1), r_{2} = 1$

$r_{1} - r_{2} = 4 - 1 = 3$

$r_{1} + r_{2} = 4 + 1 = 5$

$C_{1} C_{2} = \sqrt{3^{2} + 4^{2}} = 5$

$⟹ C_{1} C_{2} = r_{1} + r_{2}$

It has 3 common tangents

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Similar questions

x^{2} + y^{2} - 4 x - 6 y - 3 = 0

x^{2} + y^{2} + 2 x + 2 y + 1 = 0

x^{2} + y^{2} + 6 x + 4 y + 4 = 0

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x^{2} + y^{2} + 4 x - 6 y + 4 = 0

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C_{1} \equiv x^{2} + y^{2} - 4 = 0

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C_{3} \equiv x^{2} + y^{2} - 8 x - 2 y + 16 = 0

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