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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
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+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
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Like I said before, you should sign up! I can easily recognize you anywhere. So just sign up! You're really good!

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

MathHelping  Jul 10, 2022
 #2
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+1

Nice Guest, very good explanation!wink

 Jul 10, 2022
 #3
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+1

Thank you :)!

Guest Jul 10, 2022

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