The circle centered at (2,-1) and with radius 4 intersects the circle centered at (2,6) and with radius sqrt{11} at two points A and B. Find (AB)^2.
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! :)
