Find the area of the circle inscribed in the smaller part of the sector

Question

A sector with a radius $R$ and the angle which is equal to $90^{\circ}$ ($EAB $ in the picture) has been divided in two parts with an arc of the same radius. How can I find the area of the circle which is inscribed in the smaller part of the sector?

It is obvious that I need to find the radius of the circle to find the area of the square. Let it be $r$. Then the line segment $EF$ equals $R+r$. Now let $K$ be the intersection point of the radius $AE$ and the line parallel to radius $AB$ which comes throught the point $F$ (center of the circle the radius of which I need to find). From here we can compose a Pythagorean theorem since $CE=R-r$. We can find $CF$ from it. I am stuck at this point, what shall I do next? Or maybe there is a simpler solution of this problem?

Answer 1

You can apply Pythagoras' theorem also to triangle $ACF$: $$ CF^2=AF^2-AC^2=EF^2-EC^2, $$ that is: $$ (R-r)^2-r^2=(R+r)^2-(R-r)^2. $$

Answer 2

The radius of the wanted circle can be found through the Pythagorean theorem as pointed out by Aretino. A construction through straightedge and compass can be devised through the following lines:

1. By circle inversion, it is enough to find the inscribed circle in a region delimited by two lines and a circle;
2. The previous problem can be easily solved by intersecting an angle bisector and a parabola.