lf the number of common tangents of the circles $x^{2} + y^{2} + 8 x + 6 y + 21 = 0, x^{2} + y^{2} + 2 y - 15 = 0$ are 2, then the point of their intersection is

(- 4, - 3)

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(- 4, - 1)

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(8, - 5)

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(8, 5)

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Solution

The correct option is D $(- 4, - 1)$
[ From point (1) ]
$S_{1} : x^{2} + y^{2} + 8 x + 6 y + 21 = 0$
$C_{1} \equiv (- 4, - 3)$ & $r_{1} = 2$
$S_{2} : x^{2} + y^{2} + 2 y - 15 = 0$
$C_{2} \equiv (0, - 1)$ & $r_{2} = 4$
Given, $S_{1}$ & $S_{2}$ has two common tangents.
This means $S_{1}$ & $S_{2}$ intersecting each other.
equation of chord $S_{1} - S_{2} = 8 x + 4 y + 36 = 0$
$2 x + y + 9 = 0$
Put $y = - (2 x + 9)$ in $S_{2}$ .
$∴ x^{2} + (- 2 x - 9)^{2} - 2 (2 x + 9) - 15 = 0$
$5 x^{2} + 36 x + 81 - 4 x - 33 = 0$
$x = \frac{- 32 \pm \sqrt{64}}{10}$
$x = \frac{- 32 \pm 8}{10}$
$x = - 4, \frac{- 12}{5}$
Substituting in the equation of common chord we get $y = - 1$ when $x = - 4$ , and $y = - 21 / 5$ when $x = - 12 / 5$

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