The circle centered at (2,-1) and with radius 4 intersects the circle centered at (2,5) and with radius 2*sqrt(3) at two points A and B. Find AB^2.
+2668 The equation of the first circle is \((x - 2)^2 + (y + 1)^2 = 16\), and the equation of the second circle is \((x - 2)^2 + (y - 5)^2 = 12\)
Subtracting the second equation from the first gives us \((y + 1)^2 - (y - 5)^2 = 4\)
Solving for y, we find that \(y = -{7 \over 3}\).
Now, plug this into the first equation (either equation works), and solve for x: \((x-2)^2 + ({-4 \over 3})^2 = 16\)
With your x values, calculate the distance between them, and square it...
Good luck!
