Intersection of two circles

Introduction

Given (see figure 1 and figure 2): $M_1=\left(x_1,y_1\right)$ , $M_2=\left(x_2,y_2\right)$ , $r_1$ and $r_2$.

From Pythagoras:

$$d=\sqrt{{\left(x_2-<!--\; -\; -->x_1\right)}^2+{\left(y_2-<!--\; -\; -->y_1\right)}^2}$$

• If $d>r_1+r_2$, then no solutions; one circle completely outside the other.
• If , then no solutions; one circle completely inside the other.
• If d = 0 and $r_1=r_2$, then infinite number of solutions; circles are coincident.

Pythagoras again:

$$\begin{array}{lll}\left[1\right]& △<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->M_1<!--\; INVISIBLE\; SEPARATOR\; -->M<!--\; INVISIBLE\; SEPARATOR\; -->P& \text{:}& a^2+h^2={r_1}^2△<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->M_2<!--\; INVISIBLE\; SEPARATOR\; -->M<!--\; INVISIBLE\; SEPARATOR\; -->P& \text{:}& b^2+h^2={r_2}^2\end{array}$$

Subtract both equations:

$$\begin{array}{rcl}{a}^2-<!--\; -\; -->b^2& =& {r_1}^2-<!--\; -\; -->{r_2}^2\\ a^2-<!--\; -\; -->{\left(d-<!--\; -\; -->a\right)}^2& =& {r_1}^2-<!--\; -\; -->{r_2}^2\\ a^2-<!--\; -\; -->d^2+2<!--\; INVISIBLE\; TIMES\; -->d<!--\; INVISIBLE\; TIMES\; -->a-<!--\; -\; -->a^2& =& {r_1}^2-<!--\; -\; -->{r_2}^2\\ -<!--\; -\; -->d^2+2<!--\; INVISIBLE\; TIMES\; -->d<!--\; INVISIBLE\; TIMES\; -->a& =& {r_1}^2-<!--\; -\; -->{r_2}^2\end{array}$$

Hence:

$$a=\frac{{r_1}^2-<!--\; -\; -->{r_2}^2+d^2}{2<!--\; INVISIBLE\; TIMES\; -->d}$$

How much is a when the two circles touch at one point, i.e. $d=r_1+r_2$ ?

Then substituting a in [1] yields:

$$h=\sqrt{{r_1}^2-<!--\; -\; -->a^2}$$

Similar triangles in figure 3:

$△<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->W<!--\; INVISIBLE\; SEPARATOR\; -->M_1<!--\; INVISIBLE\; SEPARATOR\; -->M_2\sim <!--\; ~\; -->△<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->U<!--\; INVISIBLE\; SEPARATOR\; -->M<!--\; INVISIBLE\; SEPARATOR\; -->P$ :

$$\frac{x_m-<!--\; -\; -->x_p}{y_2-<!--\; -\; -->y_1}=\frac{h}{d}$$ Or for the other intersection point (not visible in figure 3): $$\frac{x_p-<!--\; -\; -->x_m}{y_2-<!--\; -\; -->y_1}=\frac{h}{d}$$

Hence:

$$x_p=x_m\pm <!--\; \pm \; -->\frac{h}{d}<!--\; INVISIBLE\; TIMES\; -->\left(y_2-<!--\; -\; -->y_1\right)$$

$△<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->W<!--\; INVISIBLE\; SEPARATOR\; -->M_1<!--\; INVISIBLE\; SEPARATOR\; -->M_2\sim <!--\; ~\; -->△<!--\; △\; --><!--\; INVISIBLE\; SEPARATOR\; -->V<!--\; INVISIBLE\; SEPARATOR\; -->M_1<!--\; INVISIBLE\; SEPARATOR\; -->M$ :

$$\frac{x_m-<!--\; -\; -->x_1}{a}=\frac{x_2-<!--\; -\; -->x_1}{d}$$ ⇔ $$x_m=\frac{a}{d}<!--\; INVISIBLE\; TIMES\; -->\left(x_2-<!--\; -\; -->x_1\right)+x_1$$

And substituting x_m in the previous equation for x_p results in:

$$x_p=\frac{a}{d}<!--\; INVISIBLE\; TIMES\; -->\left(x_2-<!--\; -\; -->x_1\right)+x_1\pm <!--\; \pm \; -->\frac{h}{d}<!--\; INVISIBLE\; TIMES\; -->\left(y_2-<!--\; -\; -->y_1\right)$$

And similar for the y-coordinates:

$$y_p=\frac{a}{d}<!--\; INVISIBLE\; TIMES\; -->\left(y_2-<!--\; -\; -->y_1\right)+y_1\mp <!--\; MINUS-OR-PLUS\; SIGN\; -->\frac{h}{d}<!--\; INVISIBLE\; TIMES\; -->\left(x_2-<!--\; -\; -->x_1\right)$$

So, this yields two xy-coordinates \left(x_p,y_p\right), one for each intersection point. If both circles touch in one point (h = 0), then {\left(x_p,y_p\right)}_1={\left(x_p,y_p\right)}_2.