If a circle C, whose radius is 3, touches externally the circle x^2+y^2+2x-4y-4=0 at the p…

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

Accepted Answer

Given circle is :

x2 + y2 + 2x - 4y -4 = 0

$$\therefore\,\,\,$$ its center is (- 1, 2) and radius is 3 units.

Let A = (x, y) be the center of the circle C

$$ \therefore $$ $\,\,\,$${{x - 1} \over 2}$= 2$ \Rightarrow $x = 5 and${{y + 2} \over 2}$= 2$ \Rightarrow $$ y = 2

So the center of C is  (5, 2)  and its radius is 3

$$\therefore\,\,\,$Equation of center C is :

x2 + y2$-$10x$-$4y + 20 = 0$\therefore\,\,\,$$ The length of the intercept it cuts on the x-axis

= 2$$\sqrt {{g^2} - c}  = 2\sqrt {25 - 20}  = 2\sqrt 5 $

Solutions · 1