Show that the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ touch each other. Also find the point of contact and common tangent at this point of contact.

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Solution

The equations of circles are
$x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$
Centres are $C_{1} (2, 3), C_{2} (- 3, - 9)$
$r_{1} = \sqrt{4 + 9 + 12} = 5$
$r_{2} = \sqrt{9 + 81 - 26} = 8$
$C_{1} C_{2} = \sqrt{(2 + 3)^{2} + (3 + 9)^{2}}$
$= \sqrt{25 + 144}$
$= 13 = r_{1} + r_{2}$
Therefore, circles touch externally.
Equation of common tangent is $S_{1} - S_{2} = 0$ .
$- 10 x - 24 y - 38 = 0$
$5 x + 12 y + 19 = 0$
The point of contact $P$ divides $C_{1} C_{2}$ in the ratio $5 : 8$ .
Co-ordinates of $P$ are:
$(\frac{5 (- 3) + 8.2}{5 + 8}, \frac{5 (- 9) + 8.3}{5 + 8}) = (\frac{1}{13}, \frac{- 21}{3})$

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