Question

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

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

Answer 1

Answer: (B)

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

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.

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

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

$\therefore$ Locus of (h, k) is

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