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Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 8, find r^2, where r is the radius of the circle.

 

 Apr 13, 2022

Best Answer 

 #1
avatar+578 
+2

\(8\cdot \sqrt{20^2+r^2}=\left(r-20\right)\left(r+20\right)\)

\(64\cdot 400+49r^2=r^4-800r^2+400^2=0\)

\(r^2= \left(\sqrt{\frac{849+\sqrt{183201}}{2}}\right)^2\)

\(r^2= approx.638.5, r = approx.25.3\)

.
 Apr 13, 2022
 #1
avatar+578 
+2
Best Answer

\(8\cdot \sqrt{20^2+r^2}=\left(r-20\right)\left(r+20\right)\)

\(64\cdot 400+49r^2=r^4-800r^2+400^2=0\)

\(r^2= \left(\sqrt{\frac{849+\sqrt{183201}}{2}}\right)^2\)

\(r^2= approx.638.5, r = approx.25.3\)

Vinculum Apr 13, 2022

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