Question

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

• A. $x ^ { 2 } + y ^ { 2 } + 3 x - 6 y + 5 = 0$
• B. $x ^ { 2 } + y ^ { 2 } + 3 x - 6 y - 31 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 3 x - 6 y + \frac { 29 } { 4 } = 0$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

Let (h, k)be the centre of the circle which rolls on the outside of the given circle. Centre of the given circle is $\left( \frac { - 3 } { 2 } , 3 \right)$ and its radius $= \sqrt { \frac { 9 } { 4 } + 9 + 9 } = \frac { 9 } { 2 }$.

Clearly, (h, k) is always at a distance equal to the sum $\left( \frac { 9 } { 2 } + 2 \right)$ $= \frac { 13 } { 2 }$ of the radii of two circles from $\left( - \frac { 3 } { 2 } , 3 \right)$ . Therefore $\left( h + \frac { 3 } { 2 } \right) ^ { 2 } + ( k - 3 ) ^ { 2 } = \left( \frac { 13 } { 2 } \right) ^ { 2 }$

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

$\Rightarrow$Hence locus of (h, k) is $x ^ { 2 } + y ^ { 2 } + 3 x - 6 y - 31 = 0$.