Question

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

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

Answer 1

The correct option is D $2\sqrt{5}$

Given: Radius of circle is $3$

Radius of other circle $=\sqrt{g^2+f^2-c}$ $=\sqrt{(-1{)}^2+2^2-(-4)}=3$

Now, circle $C$ is touching externally other circle at point $(2,2)$

Therefore, $(2,2)$ is midpoint of two centres of the circles as radius are equal.

Let the coordinate of centre of circle $C$ be $(p,q)$.

Centre of other circle is $(-1,2)$

$\Rightarrow \frac{p-1}{2}=2$ and $\frac{q+2}{2}=2$

$\Rightarrow p=5$ and $q=2$

Equation of circle $C$ is

$(x-5{)}^2+(y-2{)}^2=3^2$

$\Rightarrow x^2+y^2-10x-4y+20=0$

x-intercept $=2\sqrt{g^2-c}$

Here, $g=(-5)$ and $c=20$

x-intercept $=2\sqrt{(-5{)}^2-20}$

$=2\sqrt{5}$