The locus of the centre of a circle of radius | vedclass

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Question

(B) Let $(h, k)$ be the centre of the circle of radius $r_2 = 2$ that rolls on the outside of the given circle.The given circle is ${x^2} + {y^2} + 3x

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $(h, k)$ be the centre of the circle of radius $r_2 = 2$ that rolls on the outside of the given circle.The given circle is ${x^2} + {y^2} + 3x - 6y - 9 = 0$.Its centre $C_1$ is $\left( -\frac{3}{2}, 3 \right)$ and its radius $r_1 = \sqrt{\left( \frac{3}{2} \right)^2 + (-3)^2 - (-9)} = \sqrt{\frac{9}{4} + 9 + 9} = \sqrt{\frac{9 + 72}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.Since the circle rolls on the outside,the distance between the centres $(h, k)$ and $\left( -\frac{3}{2}, 3 \right)$ is equal to the sum of the radii,$r_1 + r_2 = \frac{9}{2} + 2 = \frac{13}{2}$.Thus,the locus satisfies: $\left( h + \frac{3}{2} \right)^2 + (k - 3)^2 = \left( \frac{13}{2} \right)^2$.Expanding this: $h^2 + 3h + \frac{9}{4} + k^2 - 6k + 9 = \frac{169}{4}$.$h^2 + k^2 + 3h - 6k + \frac{9 + 36 - 169}{4} = 0$.$h^2 + k^2 + 3h - 6k - \frac{124}{4} = 0$.$h^2 + k^2 + 3h - 6k - 31 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is ${x^2} + {y^2} + 3x - 6y - 31 = 0$.

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

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