The number of common tangents to circle $x^{2} + y^{2} + 2 x + 8 y - 23 = 0$ and $x^{2} + y^{2} - 4 x - 10 y + 9 = 0$ , is.

1

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3

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2

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None of these

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Solution

The correct option is B $2$
$S 1 : x^{2} + y^{2} + 2 x + 8 y - 23 = 0$

$C_{1} = (- 1, - 4), r_{1}^{2} = \sqrt{(- 1)^{2} + (- 4)^{2} + 23} = 2 \sqrt{10}$

$S 2 : x^{2} + y^{2} - 4 x - 10 y + 9 = 0$

$C_{2} = (2, 5), r_{2}^{2} = \sqrt{(2)^{2} + (5)^{2} - 9} = 2 \sqrt{5}$

$C_{1} C_{2} = \sqrt{3^{2} + 9^{2}} = \sqrt{90} = 3 \sqrt{10}$

$r_{1} + r_{2} = 2 \sqrt{10} + 2 \sqrt{5} = \sqrt{10} (2 + \sqrt{2})$

$3 \sqrt{1} 0 < (2 + \sqrt{2}) \sqrt{10}$

$⟹ C_{1} C_{2} < r_{1} + r_{2}$

$| r_{1} - r_{2} | = (2 - \sqrt{2}) \sqrt{10}$

$3 \sqrt{10} > (2 - \sqrt{2}) \sqrt{10}$

$| r_{1} - r_{2} | < C_{1} C_{2} < r_{1} + r_{2}$

Only 2 common tangents.

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