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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.

Jul 26, 2022

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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!

Jul 26, 2022
edited by BuilderBoi  Jul 26, 2022