We will use the following two facts:
1) Since ABCO is a rectangle, the two diagonals AC and OB are the same length
2) Since O is the center of the circle and B is a point on the circumference, OB is a radius of the circle.
We can compute the length of AC, since we know the length of OA and OC: in fact, OAC is a right triangle of which we know the two legs. So, we have
[tex] AC = \sqrt{OA^2+OC^2} = \sqrt{5^2+6^2} = \sqrt{25+36} = \sqrt{61} [/tex]
Invoking the point (1), we can deduce that [tex] OB=AC = \sqrt{61} [/tex]
The area of a circle is given by the following formula:
[tex] A = \pi r^2 [/tex]
And by point (2), we know that OB is a radius, so we have
[tex] A = \pi (OB)^2 = \pi (\sqrt{61})^2 = 61\pi [/tex]
