$S : x^{2} + y^{2} - 16 y + 32 = 0$ and $K : x^{2} + y^{2} - 18 x + 2 y + 32 = 0$ are two circles. Show that the circles touch externally.

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Solution

$S : x^{2} + y^{2} - 16 y + 32 = 0$
centre = $(0, 8) = c_{1}$
radius = $\sqrt{0^{2} + 8^{2} - 32} = \sqrt{32} = 4 \sqrt{2}$

$K : x^{2} + y^{2} - 18 x + 2 y + 32 = 0$
centre = $(9, - 1) = c_{2}$
radius = $\sqrt{9^{2} + (- 1)^{2} - 32} = \sqrt{50} = 5 \sqrt{2}$

$r_{1} + r_{2} = 4 \sqrt{2} + 5 \sqrt{2} = 9 \sqrt{2}$
Distance between centres = $\sqrt{(9 - 0)^{2} + (- 1 - 8)^{2}} = \sqrt{162} = 9 \sqrt{2}$
Thus, the circles touch externally, as $r_{1} + r_{2} = | c_{1} c_{2} |$ .

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