Question

If a circle $C$, whose radius is $4$, touches the circle $x^2+y^2+4x-6y-3=0$ at point $(2,3)$ externally, then the length of the intercept cut by the circle $C$ on the $x-$axis is equal to

• A. $2\sqrt{7}$
• B. $2\sqrt{5}$
• C. $2\sqrt{3}$
• D. $2\sqrt{8}$

Answer 1

The correct option is A $2\sqrt{7}$

For the circle,
$x^2+y^2+4x-6y-3=0$
Centre $\equiv (-2,3)$
Radius $=\sqrt{4+9+3}=4$


This circle touches the circle $C$ at point $(2,3)$ externally.
Thus, $(2,3)$ divides the segment joining two centres in the ratio $4:4$ or $1:1$
Using section formula
$\frac{x_1-2}{2}=2\Rightarrow x_1=6\frac{y_1+3}{2}=3\Rightarrow y_1=3$
Centre of circle $C$ is $(6,3)$
Equation of circle $C$:
$(x-6{)}^2+(y-3{)}^2=16$
Putting $y=0$
$(x-6{)}^2=7\Rightarrow x=6\pm \sqrt{7}$

Length of $x-$intercept
$6+\sqrt{7}-6+\sqrt{7}=2\sqrt{7}$