Shows the largest quarter circle within a circle.

Base is the circle of given radius ${\displaystyle r_{0}}$ around point ${\displaystyle S_{0}}$
Inscribed is the largest possible quarter circle.

In order to find radius ${\displaystyle r_{1}}$ of the quarter circle, the following reasoning is used: Since point ${\displaystyle D}$ is the center of the circle we have:
${\displaystyle |AS_{0}|=|BS_{0}|=|CS_{0}|=r_{0}}$

The points ${\displaystyle A}$,${\displaystyle B}$ and ${\displaystyle C}$ form a rectangular, isosceles triangle with:
${\displaystyle |AB|=|AC|=r_{1}}$ and ${\displaystyle |BC|=2\cdot r_{0}}$

Applying the Pythagorean theorem on ${\displaystyle \triangle ABC}$ gives:
${\displaystyle |AB|^{2}+|AC|^{2}=|CB|^{2}}$
${\displaystyle \Rightarrow r_{1}^{2}+r_{1}^{2}=(2\cdot r_{0})^{2}}$
${\displaystyle \Rightarrow 2\cdot r_{1}^{2}=4\cdot r_{0}^{2}}$
${\displaystyle \Rightarrow r_{1}^{2}=2\cdot r_{0}^{2}}$
${\displaystyle \Rightarrow r_{1}={\sqrt {2\cdot r_{0}^{2}}}}$
${\displaystyle \Rightarrow r_{1}={\sqrt {2}}\cdot r_{0}}$

0) The radius of the base circle ${\displaystyle =r_{0}}$
1) Radius of the quarter circle ${\displaystyle r_{1}={\sqrt {2}}\cdot r_{0}}$

0) Perimeter of base circle ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}}$
1) Perimeter of the quarter circle ${\displaystyle P_{1}=2\cdot r_{1}+{\frac {\pi \cdot r_{1}}{2}}=(2+{\frac {\pi }{2}})\cdot r_{1}=(2+{\frac {\pi }{2}})\cdot {\sqrt {2}}\cdot r_{0}={\frac {4+\pi }{2}}\cdot {\sqrt {2}}\cdot r_{0}=(4+\pi )\cdot {\frac {\sqrt {2}}{2}}\cdot r_{0}=(4+\pi )\cdot {\frac {1}{\sqrt {2}}}\cdot r_{0}}$
${\displaystyle \Rightarrow P_{1}={\frac {4+\pi }{\sqrt {2}}}\cdot r_{0}}$

0) Area of the base circle ${\displaystyle A_{0}=\pi \cdot r_{0}^{2}}$
1) Area of the inscribed quarter circle ${\displaystyle A_{1}={\frac {\pi \cdot r_{1}^{2}}{4}}={\frac {\pi }{4}}\cdot ({\sqrt {2}}\cdot r_{0})^{2}={\frac {\pi }{4}}\cdot 2\cdot r_{0}^{2}={\frac {\pi }{2}}\cdot r_{0}^{2}={\frac {1}{2}}\cdot A_{0}}$

Centroid positions are measured from the centroid point of the base shape
0) Centroid positions of the base square: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed quarter circle:
${\displaystyle x_{1}={\frac {4\cdot r_{1}}{3\cdot \pi }}-\cdot r_{0}\cdot cos(45)={\frac {4\cdot {\sqrt {2}}\cdot r_{0}}{3\cdot \pi }}-\cdot r_{0}\cdot {\frac {1}{\sqrt {2}}}=\left({\frac {4\cdot {\sqrt {2}}}{3\cdot \pi }}-{\frac {1}{\sqrt {2}}}\right)\cdot r_{0}={\frac {4\cdot 2-3\cdot \pi }{3\cdot \pi \cdot {\sqrt {2}}}}\cdot r_{0}={\frac {8-3\pi }{3\pi {\sqrt {2}}}}\cdot r_{0}}$

${\displaystyle y_{1}=x_{1}={\frac {8-3\pi }{3\pi {\sqrt {2}}}}\cdot r_{0}}$

In the normalised case the area of the base is set to 1.
${\displaystyle A_{0}=1\quad \Rightarrow \pi \cdot r_{0}^{2}=1\quad \Rightarrow r_{0}^{2}={\frac {1}{\pi }}\quad \Rightarrow r_{0}={\frac {1}{\sqrt {\pi }}}}$

0) Radius of the base circle ${\displaystyle r_{0}={\frac {1}{\sqrt {\pi }}}=0.56...}$
1) Radius of the inscribed quarter circle ${\displaystyle r_{1}={\sqrt {2}}\cdot r_{0}={\sqrt {2}}\cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\frac {2}{\pi }}}=0.79...}$

0) Perimeter of base square ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}=2\cdot \pi \cdot {\frac {1}{\sqrt {\pi }}}=2\cdot {\sqrt {\pi }}=3.5449077...}$
1) Perimeter of the inscribed quarter circle ${\displaystyle P_{1}={\frac {4+\pi }{\sqrt {2}}}\cdot r_{0}={\frac {4+\pi }{\sqrt {2}}}\cdot {\frac {1}{\sqrt {\pi }}}={\frac {4+\pi }{\sqrt {2\cdot \pi }}}=2.8490832...}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}=6.3939909...}$

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the inscribed quarter circle ${\displaystyle A_{1}={\frac {\pi }{2}}\cdot r_{0}^{2}={\frac {\pi }{2}}\cdot ({\frac {1}{\sqrt {\pi }}})^{2}={\frac {\pi }{2}}\cdot {\frac {1}{\pi }}=0.5}$

Centroid positions are measured from the centroid point of the base shape.
0) Centroid positions of the base square: ${\displaystyle S_{0}:\quad x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed quarter circle:${\displaystyle S_{1}:\quad x_{1}={\frac {8-3\pi }{3\pi {\sqrt {2}}}}\cdot r_{0}={\frac {8-3\pi }{3\pi {\sqrt {2}}}}\cdot {\frac {1}{\sqrt {\pi }}}=-0.0603095...\quad y_{1}=x_{1}=-0.0603095...}$

The distance between the centroid of the base element and the centroid of the quarter circle is:
${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.0852905...}$

Apart of the base element there is only one shape allocated. Therefore the integer part of the identifying number is 1.
The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case.

${\displaystyle decimalpart(6.3939910...+0.0852906...)=decimalpart(6.4792816...)=.4792816}$

So the identifying number is: ${\displaystyle 1.4792816}$

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