# Three Kissing Circles

If three circles with radii $$x,y$$, and $$z$$ make tangents on each other as shown, what will be the area of the figure(magenta colored) formed between them in terms of $$x,y$$, and $$z$$? I am posting my attempt below $$\frac { 1 }{ 2 } \left\{ \left( { x }^{ 2 }+xy+xz+yz \right) \ast \sin { \left( \cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) } \right) } -\quad { x }^{ 2 }\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }+xy+xz-yz }{ { x }^{ 2 }+xy+xz+yz } \right) -\quad { y }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { y }^{ 2 }+xy+yz-xz }{ { y }^{ 2 }+xy+yz+xz } \right) -\quad { z }^{ 2 } } \cos ^{ -1 }{ \left( \frac { { z }^{ 2 }+xz+yz-xy }{ { z }^{ 2 }+xz+yz+xy } \right) } \right\}$$

Note by Bilal Akmal
6 years ago

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Can you please elaborate on how you produced the formula.

- 6 years ago

Form a triangle with the circles' origins as vertices.This should be equal to the magenta area plus some sectors of each circle. Minus the sectors and you get the above equation(angles assumed in radian.)

- 6 years ago

Area of triangle - sum of the areas of three sectors

- 6 years ago

Yeah, it works out fine. I get the same result, although not exactly with the same expression.

- 6 years ago

"Update on the formula to calculate the magenta area": This one works too :)(angles are again assumed in radian) $\sqrt { { x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 } } -\frac { { x }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2y } \right) } +\sin ^{ -1 }{ \left( \frac { y+z }{ 2x+2z } \right) } \right] -\frac { { y }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+z }{ 2x+2y } \right) +\sin ^{ -1 }{ \left( \frac { x+z }{ 2y+2z } \right) } } \right] -\frac { { z }^{ 2 } }{ 2 } \left[ \sin ^{ -1 }{ \left( \frac { x+y }{ 2x+2z } \right) +\sin ^{ -1 }{ \left( \frac { x+y }{ 2y+2z } \right) } } \right]$

- 5 years, 12 months ago