If the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $5 (x^{2} + y^{2}) - 8 x - 14 y - 32 = 0$ touches each other, then the point of contact of the circles is

(1, - 1)

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(- 1, 1)

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(1, 2)

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(- 1, - 1)

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Solution

The correct option is D $(- 1, - 1)$
$S_{1} : x^{2} + y^{2} - 4 x - 6 y - 12 = 0 \Rightarrow C_{1} \equiv (2, 3)$
$S_{2} : x^{2} + y^{2} - \frac{8}{5} x - \frac{14}{5} y - \frac{32}{5} = 0 \Rightarrow C_{2} \equiv (\frac{4}{5}, \frac{7}{5})$
Equation of common tangent of two circle touching each other, is
$S_{2} - S_{1} = 0$
$\frac{12 x}{5} + \frac{16}{5} y + \frac{28}{5} = 0$
$\Rightarrow 3 x + 4 y + 7 = 0$ ...(1)
Equation of line passing through their centres
$y - 3 = \frac{4}{3} (x - 2)$
$3 y - 4 x = 1$ ...(2)
From (1) & (2),
$25 y = - 25$
$y = - 1, x = - 1$
$∴$ The point of contact of the circles is $(- 1, - 1)$

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