Circle In A Quarter

r the radius of small circle and R of the quarter-circle. The distance between circle centers is $\sqrt2$ r. Distance from center of small to the point of tangency=r. But sum of these two distances =R. That is R=( $\sqrt2+1)$ r.
Required ratio $=\dfrac{\pi*r^2}{\frac{\pi*R^2} 4}=\dfrac{r^2}{\frac{(\sqrt2+1)^2*r^2} 4}=0.686$

$R$ = Radius of the quarter circle

$r$ = radius of inner circle

If you draw a line in 45 degrees from the right corner at the bottom of the quartercircle through the midpoint of the full circle, you get a distance on this line from the right corner(=starting point) to the point, where the line first touches the full circle.

This distance is $s$ .

Draw it. Then everything gets clear.

Two Conditions:

1) $s = R - 2r$

2) $(r+s)^2 = 2r^2$

Solving the system results in:

$r^2+2Rr+R^2=0$

$\Rightarrow r = 0.414R$ .

Ratio of the two areas:

$\text{Area of circle}/ \text{ Area of sector} =$
$\pi(0.414R)^2/(\pi/4 \cdot R^2)=$
$0.686$

Join the origin of the quadrant to center of inscribed circle extended to the point of tangency of circle and quadrant (P) . Let $O_1$ be center of inscribed circle and the origin of quadrant $O$ .

Let $r , R$ be radius of radius of inscribed circle and quadrant. Now use properties of tangent to show that:

$r= R(\sqrt{2} -1)$

Hence the ratio $\zeta$ = $4(3 -2\sqrt{2} )$