File:FS QVC2 dia.png

Shows the second largest circle in a square that contains also the largest quarter circle

Base is the square ${\displaystyle [ABCD]}$ of side length s.
Inscribed is the largest possible quarter circle of radius ${\displaystyle r_{1}}$ around point ${\displaystyle A}$.
Added is the second largest circle with radius ${\displaystyle r_{2}}$ around point ${\displaystyle S_{2}}$.

0) The side length of the base square: ${\displaystyle |AB|=|BC|=|CD|=|AD|=s}$
1) Radius of the quarter circle: ${\displaystyle r_{1}=|AB|=s}$
2) Radius of the circle around point ${\displaystyle S_{2}}$: ${\displaystyle r_{2}=(3-2{\sqrt {2}})\cdot s}$. The calculation is shown under Calculation 1 below.

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot s}$
1) Perimeter of the quarter circle: ${\displaystyle P_{1}={\frac {4+\pi }{2}}\cdot s}$
2) Perimeter of the circle: ${\displaystyle P_{2}=2\pi r_{2}=2\pi \cdot (3-2{\sqrt {2}})\cdot s}$

0) Area of the base square: ${\displaystyle A_{0}=s^{2}}$
1) Area of the inscribed quarter circle: ${\displaystyle A_{1}={\frac {\pi }{4}}\cdot s^{2}}$
2) Area of the circle: ${\displaystyle A_{2}=\pi r_{2}^{2}=\pi \cdot (3-2{\sqrt {2}})^{2}\cdot s^{2}=\pi \cdot (9-2\cdot 3\cdot 2{\sqrt {2}}+4\cdot 2)\cdot s^{2}=\pi \cdot (17-12{\sqrt {2}})\cdot s^{2}}$

Centroid positions are measured from the centroid position 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: ${\displaystyle x_{1}={\frac {8-3\pi }{6\pi }}\cdot s\quad y_{1}={\frac {8-3\pi }{6\pi }}\cdot s}$
2) Centroid positions of the circle: ${\displaystyle x_{2}=y_{2}={\frac {4{\sqrt {2}}-5}{2}}\cdot s\quad }$The calculation is shown under Calculation 2 below.

In the normalised case the area of the base is set to 1.
${\displaystyle ||ABCD||=1\Rightarrow s^{2}=1\Leftrightarrow s=1}$

0) Side length of the base square: ${\displaystyle s=1}$
1) Radius of the inscribed quarter circle: ${\displaystyle r_{1}=s=1}$
2) Radius of the circle: ${\displaystyle r_{2}=(3-2{\sqrt {2}})\approx 0.172}$

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot s=4}$
1) Perimeter of the inscribed quarter circle: ${\displaystyle P_{1}={\frac {4+\pi }{2}}\approx 3.571}$
2) Perimeter of the circle: ${\displaystyle P_{2}=2\pi \cdot (3-2{\sqrt {2}})\approx 1.078}$

S) Sum of perimeters ${\displaystyle P_{s}\approx 8.649}$

0) Area of the base square: ${\displaystyle A_{0}=1}$
1) Area of the inscribed quarter circle: ${\displaystyle A_{1}={\frac {\pi \cdot r_{1}^{2}}{4}}={\frac {\pi }{4}}\quad \approx 0.785}$
2) Area of the circle: ${\displaystyle A_{2}=\pi \cdot (17-12{\sqrt {2}})\quad \approx 0.093}$

Centroid positions are measured from the centroid of the base square

0) Centroid positions of the base square: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed quarter circle: ${\displaystyle x_{1}={\frac {8-3\pi }{6\pi }}=-0.0755868...\quad y_{1}=x_{1}=-0.0755868...}$
2) Centroid positions of the circle: ${\displaystyle x_{2}={\frac {4{\sqrt {2}}-5}{2}}=0.3284271...\quad y_{2}=x_{2}=0.3284271...}$

The distances between the centroids are:
${\displaystyle d01={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.1068959...}$
${\displaystyle d02={\sqrt {(x_{0}-x_{2})^{2}+(y_{0}-y_{2})^{2}}}=0.4644661...}$
${\displaystyle d12={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}=0.5713620...}$
${\displaystyle Sum=1.1427240...}$

Apart of the base element there are two shapes allocated. Therefore the integer part of the identifying number is 2.
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.

${\displaystyle decimalpart(8.6488205...+1.1427240...)=decimalpart(9.7915445...)=.7915445}$

So the identifying number is: ${\displaystyle 2.7915445}$

(1) ${\displaystyle \quad |AC|=|AH|+|HS_{2}|+|S_{2}C|={\sqrt {2}}\cdot s}$, since [ABCD] is a square and [AC] is the diagonal.

(2) ${\displaystyle \quad |AH|=r_{1}=s}$, since the segment [AH] is the radius of the quarter circle.

(3) ${\displaystyle \quad |HS_{2}|=r_{2}}$, since the segment is the radius of the circle.

The vertices ${\displaystyle [S_{2}FCE]}$ form a square of side length ${\displaystyle r_{2}}$, so:
(4)${\displaystyle \quad |S_{2}C|={\sqrt {2}}\cdot r_{2}}$

Substituting from (2), (3) and (4) into (1) gives:
${\displaystyle |AH|+|HS_{2}|+|S_{2}C|=|AC|}$
${\displaystyle \quad \Leftrightarrow r_{1}+r_{2}+{\sqrt {2}}\cdot r_{2}={\sqrt {2}}\cdot s}$
${\displaystyle \quad \Leftrightarrow s+(1+{\sqrt {2}})\cdot r_{2}={\sqrt {2}}\cdot s}$
${\displaystyle \quad \Leftrightarrow (1+{\sqrt {2}})\cdot r_{2}=({\sqrt {2}}-1)\cdot s}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {{\sqrt {2}}-1}{1+{\sqrt {2}}}}\cdot s}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {{\sqrt {2}}-1}{1+{\sqrt {2}}}}\cdot {\frac {{\sqrt {2}}-1}{{\sqrt {2}}-1}}\cdot s}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {2-2{\sqrt {2}}+1}{{\sqrt {2}}-1+2-{\sqrt {2}}}}\cdot s}$
${\displaystyle \quad \Leftrightarrow r_{2}=(3-2{\sqrt {2}})\cdot s}$

${\displaystyle \quad x2=|S_{0}S_{2}|\cdot cos{45}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left(|S_{0}C|-|S_{2}C|\right)\cdot cos{45}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {{\sqrt {2}}\cdot s}{2}}-|S_{2}C|\right)\cdot cos{45}}$, since ${\displaystyle S_{0}}$ is the center point of the square ${\displaystyle [ABCD]}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {{\sqrt {2}}\cdot s}{2}}-{\sqrt {2}}\cdot r_{2}\right)\cdot cos{45}}$, since ${\displaystyle [S_{0}C]}$ is the diagonale of the square ${\displaystyle [S_{2}FCE]}$ with side length ${\displaystyle r_{2}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {{\sqrt {2}}\cdot s}{2}}-{\sqrt {2}}\cdot r_{2}\right)\cdot {\frac {1}{\sqrt {2}}}}$, since ${\displaystyle cos{45}={\frac {1}{\sqrt {2}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {s}{2}}-r_{2}\right)}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {s}{2}}-(3-2{\sqrt {2}})\cdot s\right)}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {1}{2}}-(3-2{\sqrt {2}})\right)\cdot s}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {1}{2}}-3+2{\sqrt {2}}\right)\cdot s}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {1-6+4{\sqrt {2}}}{2}}\right)\cdot s}$
${\displaystyle \quad \Leftrightarrow x_{2}={\frac {4{\sqrt {2}}-5}{2}}\cdot s}$

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