The circle $x^{2} + y^{2} + 6 x + 6 y = 0$ and $x^{2} + y^{2} - 12 x - 12 y = 0$

Cut orthogonally

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Touch each other internally

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Intersect in two points

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Touch each other externally

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Solution

The correct option is B Touch each other externally
Given equation of circles are
$x^{2} + y^{2} + 6 x + 6 y = 0$ ....(i)
and $x^{2} + y^{2} - 12 x - 12 y = 0$ ....(ii)
Here, $g_{1} = 3, f_{2} = 3, g_{2} = - 6$ and $f_{2} = - 6$
$∴$ Centres of circles are $C_{1} (- 3, - 3)$ and $C_{2} (6, 6)$ respectively and radii are $r_{1} = 3 \sqrt{2}$ and $r_{2} = 6 \sqrt{2}$ respectively.
Now, $C_{1} C_{2} = \sqrt{{(6 + 3)}^{2} + {(6 + 3)}^{2}}$
$= 9 \sqrt{2}$
and $r_{1} + r_{2} = 3 \sqrt{2} + 6 \sqrt{2} = 9 \sqrt{2}$
$\Rightarrow C_{1} C_{2} = r_{1} + r_{2}$
$∴$ Both circles touch each other externally.

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