Application of Power Theorem

$Solution:$

Since $AB$ is $perpendicular$ to $CD$ , it will form a right $\angle$ resulting to $Triangle (CGO)$ as a $Right$ Triangle.

Then, by using Pythagorean Theorem on $Triangle (CGO)$ , we will find that $GO = \sqrt{16-r^{2}}$

By using Power theorem,

$CG*GM$ = $AG* GB$

-> $4*3 = (r-\sqrt{16-r^{2}})*(r+\sqrt{16-r^{2}})$

-> $12 = r^{2} - (16 - r^{2})$

-> $2r^{2} = 28$

-> $r^2 = 14$

$\therefore$ , $Area (Circle O) = \pi(r^{2})$ -> $22/7 * 14$ -> $\boxed{44 sq. units}$

(This solution does not make use of the power theorem.)

Let $O$ be the center of the circle.

Let $r$ be the radius of circle $O$

Drawing the line $MD$ we can see that $\bigtriangleup$ $CMD$ is a right triangle with $\angle$ $CMD$ = $90^{\circ}$ .

$\bigtriangleup$ $COG$ is also a right triangle and shares $\angle$ $MCD$ with $\bigtriangleup$ $CMD$ , so $\bigtriangleup$ $COG$ $\sim$ $\bigtriangleup$ $CMD$ .

Thus,

$\frac {CG}{CO}$ = $\frac {CD}{CM}$

$\frac {4}{r}$ = $\frac {2r}{7}$

$2r^2$ = $28$

$r^2$ = $14$

Therefore the area of the circle is $\frac {22}{7}$ $\cdot$ $14$ = $\boxed {44}$