coordinates

The equation of the circle with centre (h,k) is: $(x-h)^2+(y-k)^2=r^2$, where r is the radius of the circle.

So:

$(x-2)^2+(y+1)^2=16$ is the equation of the first circle.                 (1)

$(x-2)^2+(y-6)^2=11$ is the equation of the second circle.           (2)

If they intersect, then we need to solve these system of equations simultaneously.

From (1):

$(x-2)^2=16-(y+1)^2$

Substitute this in (2), to get:

$16-(y+1)^2+(y-6)^2=11$   (Expand the brackets).

$16-y^2-2y-1+y^2-12y+36=11$

$-14y=-40 \implies y=\frac{20}{7}$

Next, from (1):

$(x-2)^2=16-(\frac{27}{7})^2 \implies x=\frac{\pm\sqrt{55}}{7}+2$

Hence the points are: $A(\frac{\sqrt{55}}{7}+2,\frac{20}{7}) \text{,and } B(\frac{-\sqrt{55}}{7}+2,\frac{20}{7})$

Then: $(AB)^2=((\frac{\sqrt{55}}{7}+2)-(\frac{-\sqrt{55}}{7}+2))^2+(\frac{20}{7}-\frac{20}{7})^2$  

Therefore, $(AB)^2=(\frac{2\sqrt{55}}{7})^2=\frac{220}{49}$

I hope this helps! :)

Nice Guest, very good explanation!