A rectangle $OACB$ with two axes as two sides,the origin $O$ as a vertex is drawn in which the length $OA$ is four times the width $OB$.A circle is drawn passing through the points BC and touching $OA$ at its mid-point,thus dividing the rectangle into three parts. Find the ratio of the areas of these three parts.
I found out the co-ordinates of the points $A(0,4a),B(a,0),C(a,4a)$.I found out the co-ordinates of the point where the circle touches the line $OA$ to be $M(0,2a)$.
Now, I found out the centre of the circle just to verify whether $BC$ is the diameter to be $O~'(\frac52a,2a)$. I can understand the the ratio of $AMC$ to $OMB$ is $1$ by symmetry. But, I cannot prove it. Also I have no idea how to find out the area of $BMC$. I have no clue but to apply definite integral over the equation of the circle. Help me with this. It would be better if the solution does not include definite integral.