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The largest circle inscribed in a circle of radius ${\displaystyle r_{0}}$ that contains two equally sized circles

0) The radius of the base circle: ${\displaystyle r_{0}}$
1+2) The radii of the inscribed twin circles: ${\displaystyle r_{1}={\frac {r_{0}}{2}}\quad }$
3) The radius of the third inscribed circle: ${\displaystyle r_{3}={\frac {r_{0}}{3}}\quad }$see calculation 1

0) Perimeter of base circle: ${\displaystyle P_{0}=2\pi r_{0}}$
1+2) Perimeter of the inscribed twin circles: ${\displaystyle P_{1}+P_{2}=2\cdot 2\pi r_{1}=2\cdot 2\pi {\frac {r_{0}}{2}}=2\pi r_{0}=P_{0}\quad }$
3) Perimeter of third circle: ${\displaystyle P_{3}=2\pi r_{3}=2\pi {\frac {r_{0}}{3}}={\frac {1}{3}}P_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{0}+{\frac {1}{3}}P_{0}={\frac {7}{3}}P_{0}}$

0) Area of the base circle: ${\displaystyle A_{0}=\pi \cdot r_{0}^{2}}$
1+2) Area of the inscribed twin circles: ${\displaystyle A_{1}+A_{2}=2\cdot \pi \cdot r_{1}^{2}=2\cdot \pi \cdot \left({\frac {r_{0}}{2}}\right)^{2}=2\cdot \pi \cdot \left({\frac {r_{0}^{2}}{4}}\right)={\frac {1}{2}}\cdot \pi \cdot r_{0}^{2}={\frac {1}{2}}\cdot A_{0}}$
3) Area of the third circle: ${\displaystyle A_{3}=\pi \cdot r_{3}^{2}=\pi \cdot ({\frac {r_{0}}{3}})^{2}={\frac {1}{9}}\cdot A_{0}}$

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}=\left({\frac {r_{1}}{\sqrt {2}}}+{\frac {r_{1}}{\sqrt {2}}}i\right)={\frac {r_{0}}{2{\sqrt {2}}}}+{\frac {r_{0}}{2{\sqrt {2}}}}i\quad \approx (0.354+0.354i)\cdot r_{0}}$
2) Centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=\left(-{\frac {r_{2}}{\sqrt {2}}}-{\frac {r_{2}}{\sqrt {2}}}i\right)=-{\frac {r_{0}}{2{\sqrt {2}}}}-{\frac {r_{0}}{2{\sqrt {2}}}}i\quad \approx (-0.354-0.354i)\cdot r_{0}}$
3) Centroid position of the third inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{3}=(r_{0}-r_{3})\cdot \left({\frac {1}{\sqrt {2}}}-{\frac {1}{\sqrt {2}}}i\right)=\left(r_{0}-{\frac {r_{0}}{3}}\right)\cdot \left({\frac {1}{\sqrt {2}}}-{\frac {1}{\sqrt {2}}}i\right)=\left(1-{\frac {1}{3}}\right)\cdot \left({\frac {1}{\sqrt {2}}}-{\frac {1}{\sqrt {2}}}i\right)\cdot r_{0}=\left({\frac {2}{3{\sqrt {2}}}}-{\frac {2}{3{\sqrt {2}}}}i\right)\cdot r_{0}\quad \approx (0.471-0.471i)\cdot r_{0}}$

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq =S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the ma-thematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}=(0-r_{1}i)=0-{\frac {r_{0}}{2}}i\quad =(0-0.500i)\cdot r_{0}}$
2) Orientated centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=(0+r_{1}i)=0+{\frac {r_{0}}{2}}i\quad =(0+0.500i)\cdot r_{0}}$
3) Orientated centroid position of the third inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{3}=(r_{0}-r_{3})+0i=\left(r_{0}-{\frac {r_{0}}{3}}\right)+0i=\left({\frac {2}{3}}+0i\right)\cdot r_{0}\quad \approx (0.667+0i)\cdot r_{0}}$

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

0) Radius of the base circle ${\displaystyle r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$
1+2) Radii of the inscribed circles: ${\displaystyle r_{1}={\frac {\frac {1}{\sqrt {\pi }}}{2}}={\frac {1}{2{\sqrt {\pi }}}}\quad \approx 0.282}$
3) Radius of the third circle: ${\displaystyle r_{3}={\frac {1}{3}}\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 0.188}$

0) Perimeter of base circle: ${\displaystyle P_{0}=2\pi \cdot {\frac {1}{\sqrt {\pi }}}=2{\sqrt {\pi }}\quad \approx 3.545}$
1+2) Perimeter of the inscribed circles: ${\displaystyle P_{1}+P_{2}=P_{0}=2{\sqrt {\pi }}\quad \approx 3.545}$
3) Perimeter of third circle: ${\displaystyle P_{3}=2\pi r_{3}={\frac {1}{3}}\cdot 2{\sqrt {\pi }}\quad \approx 1.182}$
S) Sum of perimeters: ${\displaystyle P_{s}={\frac {7}{3}}P_{0}={\frac {7}{3}}\cdot 2{\sqrt {\pi }}\quad \approx 8.2714513}$

0) Area of the base square is by definition ${\displaystyle A_{0}=1}$
1+2) Area of the inscribed circles: ${\displaystyle A_{1}+A_{2}={\frac {1}{2}}\cdot A_{0}={\frac {1}{2}}=0.500}$
3) Area of the third circle: ${\displaystyle A_{3}={\frac {1}{9}}\cdot A_{0}\quad \approx 0.111}$

Covered surface of the base shape: ${\displaystyle A_{c}={\frac {A_{1}+A_{2}+A_{3}}{A_{0}}}={\frac {1}{2}}+{\frac {1}{9}}={\frac {11}{18}}\quad \approx 61\%}$

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}={\frac {1}{2{\sqrt {2\pi }}}}+{\frac {1}{2{\sqrt {2\pi }}}}i\quad \approx (0.199+0.199i)}$
2) Centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=-{\frac {1}{2{\sqrt {2\pi }}}}-{\frac {1}{2{\sqrt {2\pi }}}}i\quad \approx (-0.199-0.199i)}$
3) Centroid position of the third inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{3}=\left({\frac {2}{3{\sqrt {2}}}}-{\frac {2}{3{\sqrt {2}}}}i\right)\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx (0.266-0.266i)}$

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the ma-thematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}=\left(0-{\frac {1}{2{\sqrt {\pi }}}}i\right)\quad \approx (0-0.282i)}$
2) Orientated centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=\left(0+{\frac {1}{2{\sqrt {\pi }}}}i\right)\quad \approx (0+0.282i)}$
3) Orientated centroid position of the third inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{3}=\left({\frac {2}{3}}+0i\right)\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx (0.376+0i)}$

(1) ${\displaystyle |S_{0}A|=r_{0}}$
(2) ${\displaystyle |S_{0}S_{1}|=|S_{0}S_{2}|=r_{1}={\frac {1}{2}}\cdot r_{0}}$
(3) ${\displaystyle |S_{1}S_{3}|=|S_{2}S_{3}|=r_{1}+r_{3}={\frac {1}{2}}\cdot r_{0}+r_{3}\quad }$because the circles are tangent to each other
(4) ${\displaystyle |S_{0}S_{3}|=|S_{0}A|-|AS_{3}|=r_{0}-r_{3}}$

${\displaystyle \quad |S_{0}S_{1}|^{2}+|S_{0}S_{3}|^{2}=|S_{1}S_{3}|^{2}\quad }$, applying the Pythagorean theorem on right triangle ${\displaystyle \triangle {S_{0}S_{1}S_{3}}}$
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+|S_{0}S_{3}|^{2}=|S_{1}S_{3}|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+(r_{0}-r_{3})^{2}=|S_{1}S_{3}|^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+(r_{0}-r_{3})^{2}=(r_{1}+r_{3})^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+(r_{0}^{2}-2r_{0}r_{3}+r_{3}^{2})=r_{1}^{2}+2r_{1}r_{3}+r_{3}^{2}\quad }$, applying binomial formulas
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{3}=2r_{1}r_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=2r_{1}r_{3}+2r_{0}r_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=2({\frac {1}{2}}\cdot r_{0})r_{3}+2r_{0}r_{3}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=r_{0}r_{3}+2r_{0}r_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=3r_{0}r_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {r_{0}^{2}}{3r_{0}}}=r_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{3}={\frac {r_{0}}{3}}\quad }$, rearranging

${\displaystyle \quad S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot (S_{1}+S_{2})+A_{3}\cdot S_{3}}{A_{0}+2A_{1}+A_{3}}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {A_{0}\cdot 0+A_{1}\cdot (S_{1}+S_{2})+A_{3}\cdot S_{3}}{A_{0}+2A_{1}+A_{3}}}\quad }$, since ${\displaystyle S_{0}=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {0+{\frac {A_{0}}{4}}\cdot (S_{1}+S_{2})+A_{3}\cdot S_{3}}{A_{0}+2{\frac {A_{0}}{4}}+A_{3}}}\quad }$, since ${\displaystyle A_{1}={\frac {A_{0}}{4}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {A_{0}}{4}}\cdot (S_{1}+S_{2})+{\frac {A_{0}}{9}}\cdot S_{3}}{A_{0}+2{\frac {A_{0}}{4}}+{\frac {A_{0}}{9}}}}\quad }$, since ${\displaystyle A_{3}={\frac {A_{0}}{9}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {1}{4}}\cdot (S_{1}+S_{2})+{\frac {1}{9}}\cdot S_{3}}{1+{\frac {1}{2}}+{\frac {1}{9}}}}\quad }$, shortening
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {1}{4}}\cdot (S_{1}+S_{2})+{\frac {1}{9}}\cdot S_{3}}{\frac {18+9+2}{18}}}\quad }$, shortening
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {1}{4}}\cdot 0+{\frac {1}{9}}\cdot S_{3}}{\frac {18+9+2}{18}}}\quad }$, since ${\displaystyle S_{1}=-S_{2}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {18}{29}}\cdot {\frac {1}{9}}\cdot S_{3}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {2}{29}}\cdot S_{3}\quad }$, shortening
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {2}{29}}\cdot (0-{\frac {2}{3}}i)\cdot r_{0}\quad }$, since ${\displaystyle S_{3}=(0-{\frac {2}{3}}i)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}=(0-{\frac {4}{87}}i)\cdot r_{0}\quad }$, rearranging

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