The locus of the centre of a circle of radius 2 which rolls on the outside of the circle $x^{2} + y^{2} + 3 x - 6 y - 9 = 0$ is

x^{2} + y^{2} + 3 x - 6 y + 5 = 0

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x^{2} + y^{2} + 3 x - 6 y - 31 = 0

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x^{2} + y^{2} + 3 x - 6 y + 29 = 0

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Solution

The correct option is C $x^{2} + y^{2} + 3 x - 6 y - 31 = 0$
Let [h ,k) be the centre of the circle which rolls on the outside of the given circle. The centre of the given circle is

$(- \frac{3}{2}, 3)$ and its radius = $\sqrt{[\frac{9}{4} + 9 + 9]}$ = $\frac{9}{2}$

Clearly (h, k) is always at a distance equal to the sum

$(\frac{9}{2} + 2) = \frac{13}{2}$ of the radii of two circles.

$∴$ $(h + \frac{3}{2})^{2} + (k - 3)^{2} = (\frac{13}{2})^{2}$

$∴$ $h^{2} + k^{2} + 3 h - 6 k + \frac{9}{4} + 9 - \frac{169}{4} = 0$

$∴$ Locus of (h, k) is

$x^{2} + y^{2} + 3 x - 6 y - 31 = 0.$

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