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

Answer 1

Answer: (B)

The centre and radius of circle $x^2+y^2+3x-6y-9=0$ are $C_1\left(-\frac{3}{2},3\right)$ and $r_1=\frac{9}{2}$
Let the centre and radius of required circle are $C_2(g,f)$ and $r_2=2$ .
Since, the required circle is rolled outside the given circle.
$\therefore$ $C_1{C}_2=r_1+r_2$
$\Rightarrow$ $\sqrt{{\left(g+\frac{3}{2}\right)}^2+{(f-3)}^2}=2+\frac{9}{2}$
$\Rightarrow$ $g^2+\frac{9}{4}+3g+f^2+9-6f=31$ Hence, locus of the centre is $x^2+y^2+3x-6y-31=0$