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The circle centered at (2, -1) and with radius 4 intersects the circle centered at (2, 5) and with radius sqrt(10) at two points A and B. Find (AB)^2.

 May 3, 2022
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\(Find\ \overline{AB}\ ^2\)

 

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\(+\sqrt{16-(x-2)^2}-1=-\sqrt{10-(x-2)^2}+5\\ +\sqrt{16-(x-2)^2}+\sqrt{10-(x-2)^2}=6\ |\ square\\ 16-(x-2)^2+2\cdot \sqrt{16-(x-2)^2}\cdot\sqrt{10-(x-2)^2}+10-(x-2)^2=36\\\)

\(x\in\{2-\frac{\sqrt{15}}2{},2+\frac{\sqrt{15}}2{}\}\)

\(\overline{AB}\ ^2=15\)

laugh  !

 May 3, 2022
edited by asinus  May 3, 2022
edited by asinus  May 4, 2022

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