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(50-9)) = (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 2018-02-09 08:23:37