Question

The radius of a circle with centre O is $\sqrt{50}$ cm. A and C are two points on the circle, and B is a point inside the circle. The length of AB is 6 cm, and the length of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance OB.

Accepted Answer

A. $AC = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$. Let $A = (0,0)$. $C = (6,2)$. With radius $R = \sqrt{50}$, center $O$ lies on perpendicular bisector of $AC$. Midpoint of $AC = (3,1)$. Height of $O$ above $AC$ = $\sqrt{50 - 10} = \sqrt{40}$. Coordinates of $O = (1, 7)$ or similar. Distance $OB^2 = 26$.

Distance in Geometry of Circle

Explanation

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