Question

Centre of the circle whose radius is $3$ and which touches internally the circle $x^3+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is

• A. $(\frac{7}{5},-\frac{4}{5})$
• B. $(\frac{4}{5},\frac{7}{5})$
• C. $(\frac{3}{5},\frac{4}{5})$
• D. $(\frac{7}{5},\frac{3}{5})$

Answer 1

The correct option is B $(\frac{4}{5},\frac{7}{5})$

The given circle is
$x^2+y^3-4x-6y-12=0$ ...(i)

Whose centre is $C_1(2,3)$ and radius $r_1=C_{1}A=5$

If $C_2(h,k)$ is the centre of the circle of radius $3$ which touches the circle (i) internally at the point $A(-1,-1)$, then $C_{2}a=3$ and $C_1{C}_2=C_{1}A-C_{2}A=5-3=2$.

Thus, $C_2(h,k)$ divide $C_{1}A$ in the ratio $2:3$ internally

Therefore, $h=\frac{2(-1)+3\left(2\right)}{2+3}=\frac{4}{5}$ and $K=\frac{2(-1)+3.3}{2+3}=\frac{7}{5}$

Hence, required centre is $(\frac{4}{5},\frac{7}{5})$.