Circle-ception

The two points where the circle tangentially touches the quadrant's sides must be the same distance away from the top left corner. So a right isosceles triangle can be formed.

Let the radius of the incircle be $x$ . As shown by the first diagram, the right isosceles triangle has legs $x$ and therefore hypotenuse $x\sqrt { 2 }$ . In the second diagram, we see that $x\sqrt { 2 }+x =10$ (the radius of the quadrant).

$x+x\sqrt { 2 } =10\\ x(1+\sqrt { 2 } )=10\\ x=\frac { 10 }{ 1+\sqrt { 2 } }$

As the diameter of the circle is $2x$ , it is equal to $\frac { 20 }{ 1+\sqrt { 2 } }$ , which after the denominator is rationalised comes to $20\sqrt { 2 } -20\equiv 20(\sqrt { 2 } -1)$ . So $a=20$ , $b=2$ , $c=1$ .

$\left\lfloor \frac { { 20 }^{ 2 }\times { 2 }^{ 2 }\times { 1 }^{ 2 } }{ 20+2+1 } \right\rfloor =\left\lfloor \frac { 1600 }{ 23 } \right\rfloor =\left\lfloor 69+\frac { 13 }{ 23 } \right\rfloor =\boxed { 69 }$

Assume a chord is drawn from point of the right angle R through centre of the circle and it meets the circle at point P and Q and the tangent from the point R meets at T, then RP.RQ=RT^2. If radius is x, then RT =x, RQ=10 and RP=10-2x; Hence 10*{10-2x}=x^2 or x=10{2^0.5 -1 }

$r+r\sqrt{2}=10$ , $r$ is the radius of the inscribed circle.

Solve for $r$ and multiply by 2 to find the diameter.