Each of the points P and Q lie on a circle with its center at O and having a radius of √50.
Point R is inside the circle such that:
∠PQR = 90^{o}, PQ = 6, QR = 2
Find OR
The circle: x^2 + y^2 = 50
Let PQ be a horizontal line intersecting the top half of the circle.
Q is then at (3, sqrt(509)) = (3, sqrt(41))
P is at (3, sqrt(41))
R is at (3, sqrt(41)2)
distance OR is sqrt{3^2 + [sqrt(41)2]^2}
OR = sqrt{9 + [sqrt(41)2]^2}
OR = sqrt{9 + 41+4 4sqrt(41)}
OR = sqrt{54 4sqrt(41)}
OR = 5.328

Posted by Larry
on 20180209 08:23:37 