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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.

Jul 10, 2022

#1
+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! :)

Jul 10, 2022
#4
+1158
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nerdiest  Jul 10, 2022
#5
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Thank you! Ok. I created an account :D.

MathHelping  Jul 10, 2022
#2
+49
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Nice Guest, very good explanation!

Jul 10, 2022
#3
+1

Thank you :)!

Guest Jul 10, 2022