Question

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

• A. $x^2 + y^2 + 3x - 6y - 31 = 0$
• B. $x^2 + y^2 + 3x - 6y - \frac{29}{3} = 0$
• C. $x^2 + y^2 + 3x - 6y - 45 = 0$
• D. $x^2 + y^2 + 3x - 6y + 31 = 0$

Answer 1

Answer: (C)

$x^2 + y^2 + 3x - 6y - 45 = 0$

Answer 2

Wehave, $x^2+y^2+ 3x - 6y - 9 = 0$
$\Rightarrow \left(x+ \frac{3}{2}\right)^{2} + \left(y -3\right)^{2} = \frac{81}{4}$
Hence, Iocusof the centre of the circle of radius 3 is
$\left(x+ \frac{3}{2}\right)^{2} + \left(y -3\right)^{2} = \left(\frac{9}{2} +3\right)^{2}$
$\Rightarrow x^{2} + \frac{9}{4} + 3x + y^{2} + 9-6y = \frac{225}{4}$
$\Rightarrow x^{2} + y^{2} +3x -6y - 45 = 0$