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, on the x-axis is equal to.

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

Answer 1

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

$x^2+y^2+2x–4y–4=0$

Centre $=(-1,2)$

$radius=\sqrt{1^2+2^2+4}$

$=\sqrt{9}$

$=3$

$(\frac{h-1}{2},\frac{k+2}{2})=(2,2)$

$h-1=4,$ and $k+2=8$

$\Rightarrow h=5,k=2$

$C:(x–5{)}^2+(y–2{)}^2=3^2$

$p=\frac{|2|}{\sqrt{1}}=2$

$AM=\sqrt{3^2–2^2}=\sqrt{5}$

$AB=2AB=2\sqrt{5}$.