Question

The circle $x^2+y^2=4$ cuts the circle $x^2+y^2-2x-4=0$ at the points A and B. If the circle $x^2+y^2-4x-k=0$ passes through A and B then the value of k is

• A. $-4$
• B. 0
• C. $-8$
• D. $4$

Answer 1

Answer: (D)

$4$

Centre of the circle $x^2+y^2-2x-4=0$ is $(1,0)$

And the centre of the circle $x^2+y^2=4$ is $(0,0)$ .

Centre of the circle $x^2+y^2-4x-k=0$ is $(2,0)$

To find the intersection points of $x^2+y^2-2x-4=0$ and $x^2+y^2=4$

Put $x^2+y^2=4$ in the equation of the circle $x^2+y^2-2x-4=0$, we get

$4-2x-4=0⟹x=0⟹y=\pm 2$

$x^2+y^2-2x-4=0$ and $x^2+y^2=4$ intersect at $(0,2)$ and $(0,-2)$

If the circle $x^2+y^2-4x-k=0$ passes through $(0,2)$ and $(0,-2)$, then the distance between the centre (2,0) and the above points is equal to the radius of the circle.

$\sqrt{4+k}=\sqrt{2^2+2^2}4+k=8⟹k=4$