## Radius of Circle Inscibed in a Diamond

The diagram shows a circle inscribed in a diamond of length
$2b$
and height
$2h$
.

The equation of the line OH is
$y= \frac{h}{b} x$
.
The gradient of the line BH is
$- \frac{b}{h}$
and the line passes through the point
$(b,0)$
.
Hence
$0= - \frac{b}{h} b + c \rightarrow c= \frac{b^2}{h}$

The equation of the line BH is
$y= - \frac{b}{h} x + \frac{b^2}{h}$

Solve simultaneously the equations
$y = \frac{h}{b} , \: y= - \frac{b}{h} x + \frac{b^2}{h}$

$\frac{h}{b} = - \frac{b}{h} x + \frac{b^2}{h} \rightarrow x(\frac{h}{b} + \frac{b}{h} = \frac{b^2}{h} \rightarrow x= \frac{b^2 /h}{h/b + b /h} =\frac{b^3}{h^2 +b^2}$

Then
$y= \frac{h}{b} x = \frac{h}{b} \frac{b^3}{h^2 +b^2}= \frac{b^2 h}{h^2 +b^2}$

Tthen
$r = \sqrt{(x-b)^2 + y^2 } = \sqrt{(\frac{b^3}{h^2 +b^2} -b)^2+ (\frac{b^2 h}{h^2 +b^2})^2} =\frac{hb}{h^2+b^2} \sqrt{h^2+b^2}$