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