How to calculate the intersection points of two circles

The aim of this page is to calculate the coordinates on the intersection points between two circles $ C_1 $ and $ C_2 $.

Here are our hypotheses:

center of circle $ C_1 $ is $ P_1 = (x_1, y_1)$
center of circle $ C_2 $ is $ P_2 = (x_2, y_2)$
radius of circle $ C_1 $ is $ r_1 $
radius of circle $ C_2 $ is $ r_2 $

Our goal is to calculate the coordinates of the intersection points $ P_3 $ and $ P_4 $ as a function of $ P_1 $ , $ P_2 $ , $ r_1 $ and $ r_2 $.

Intermediate steps

Here are the intermediate steps for computing the intersection points:

Calculating the distance $ d $ between circle centers
Checking cases
Calculating the length of $ a $ and $ b $
Calculation of h
Calculating the coordinates of $ P_5 $
Calculation of vectors $ \overrightarrow{P_5P_3} $ and $ \overrightarrow{P_5P_4} $
calculating the coordinates of $ P_3 $ and $ P_4 $

Distance between centers

Let's start by calculating the $ d $, the distance between the centers. By applying the Pythagorean theorem we can write:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 } \label{eq:d} $$

Checking cases

According to the values of $ d $, we now have five cases:

if $ d > r_1 + r_2 $ the circles are too far apart and do not intersect;
if $ d < | r_1 - r_2 | $ one circle is inside the other and do not intersect;
if $ d = 0 $ and $ r_1 = r_2 $ the circles are merged and there are an infinite number of points of intersection;
if $ d = r_1 + r_2 $ there is a single intersection point;
si $ d = r_1 - r_2 $ or $ d = r_2 - r_1 $ there is a single intersection point, but one circle in inside the other;
if $ d < r_1 + r_2 $ there are two intersection points.

$ d > r_1 + r_2 $	$ d < \lvert r_1 - r_2 \rvert $

$ d = 0 $ and $ r_1 = r_2 $	$ d = r_1 + r_2 $

$ d = r_1 - r_2 $ or $ d = r_2 - r_1 $	$ d < r_1 + r_2 $

Calculating a and b

To calculte the distance $a$ let's start by writing $ h $ as a function of $ a $ and $ b $ . In the right triangles $ P_1 P_5 P_3 $ the Pythagorean theorem gives:

$$ r_1^2 = h^2 + a^2 \label{eq:h1} $$

We can apply the theorem in the right triangle $ P_2 P_5 P_3 $:

$$ r_2^2 = h^2 + b^2 \label{eq:h2} $$

By substituting \eqref{eq:h2} to \eqref{eq:h1} we get the following equation:

$$ r_1^2 - r_2^2 = a^2 - b^2 \label{eq:ab} $$

Since $ d = a +b $, we can write the folowing system of two equations:

$$ \left\{ \begin{split} & r_1^2 - r_2^2 = a^2 - b^2 \\ & d = a +b \end{split} \right. $$

By injecting $ b = d-a $ in equation \eqref{eq:ab}, we obtain the following relation:

$$ r_1^2 - r_2^2 = a^2 - (d-a)^2 $$

The $ –a^2 $ terms on each side cancel out. You can then solve for $ a $:

$$ \begin{split} r_1^2 - r_2^2 &= a^2 - d^2 - a^2 + 2ad \\ r_1^2 - r_2^2 &= - d^2 + 2ad \\ 2ad & = r_1^2 - r_2^2 + d^2 \end{split}$$

We get :

$$ a = \dfrac{r_1^2 - r_2^2 + d^2 }{2d} $$

Similarly in the triangle $ P_2 P_5 P_3 $ :

$$ b = \dfrac{r_2^2 - r_1^2 + d^2 }{2d} $$

All of these values are known:

$ r_1 $ and $ r_2 $ are the radius of the circles
$ d $ has been calculated by equation \eqref{eq:d}

Calculation of h

Once $ a $ and $ b$ are know, it becomes easy to calculate the length $ h $ by apoplying the Pythagorean theorem in the right triangles $ P_1 P_5 P_3 $

$$ r_1^2 = h^2 + a^2 $$

$$ h = \sqrt{ r_1^2 - a^2 }$$

Coordinates of P5

The next step is to calculate the coordinates of $ P_5 $.

Since the vectors $ \overrightarrow{P_1P_5} $ and $ \overrightarrow{P_1P_2} $ are colinear, we can write:

$$ \overrightarrow{P_1P_5} = \dfrac{a}{d} \times \overrightarrow{P_1P_2} $$

We can deduce the coordinates of $ P_5 $:

$$ \begin{split} x_5 &= x_1 + \dfrac{a}{d} \times (x_2 - x_1) \\ y_5 &= y_1 + \dfrac{a}{d} \times (y_2 - y_1) \end{split} $$

Vectors P₅P₃ and P₅P₄

The second to last step is the calculation of vectors $ \overrightarrow{P_5P_3} $ and $ \overrightarrow{P_5P_4} $ . Let's consider the vector $ \overrightarrow{P_1P_2} $ given by the following relation:

$$ \overrightarrow{P_1P_2} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix} $$

By multiplying this vector by a rotation matrix around the z-axis, we can calculate the perpendicular vectors:

Clockwise

$$ \overrightarrow{P_1P_2}^{\bot \circlearrowright} = \begin{pmatrix} 0 && 1 \\ -1 && 0 \end{pmatrix} \times \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix} = \begin{pmatrix} y_2 - y_1 \\ x_1 - x_2 \end{pmatrix} $$

Counterclockwise

$$ \overrightarrow{P_1P_2}\ {}^{\bot \circlearrowleft} = \begin{pmatrix} 0 && -1 \\ 1 && 0 \end{pmatrix} \times \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix} = \begin{pmatrix} y_1 - y_2 \\ x_2 - x_1 \end{pmatrix} $$

We already calculated the norm of vector $ \lVert \overrightarrow{P_1P_2} \rVert = d $ and norm of vectors $ \lVert \overrightarrow{P_5P_3} \rVert = \lVert \overrightarrow{P_5P_4} \rVert = h $.

By applyting the ratio $ \dfrac{h}{d} $ to the vectors $ \overrightarrow{P_1P_2}^{\bot} $ we can deduce the expressions of vectors $ \overrightarrow{P_5P_3}^{\bot} $ and $ \overrightarrow{P_5P_4}^{\bot} $:

$$ \overrightarrow{P_5P_3}^{\bot} = \dfrac{h}{d} \times \overrightarrow{P_1P_2}^{\bot \circlearrowleft} =\begin{pmatrix} \dfrac{h(y_2 - y_1)}{d} \\ \dfrac{h(x_1 - x_2)}{d} \end{pmatrix} $$

$$ \overrightarrow{P_5P_4}^{\bot} = \dfrac{h}{d} \times \overrightarrow{P_1P_2}^{\bot \circlearrowright} =\begin{pmatrix} \dfrac{h(y_1 - y_2)}{d} \\ \dfrac{h(x_2 - x_1)}{d} \end{pmatrix} $$

Intersection points

Once the vectors $ \overrightarrow{P_5P_3} $ and $ \overrightarrow{P_5P_4} $ are known, the coordinates of $P_3$ and $P_4$ can be deduced by translating $ P_3$ from these vectors. We finally get:

$$ P_3 =\begin{pmatrix} x_5 - \dfrac{h(y_2 - y_1)}{d} \\ y_5 + \dfrac{h(x_2 - x_1)}{d} \end{pmatrix} $$

and

$$ P_4 =\begin{pmatrix} x_5 + \dfrac{h(y_2 - y_1)}{d} \\ y_5 - \dfrac{h(x_2 - x_1)}{d} \end{pmatrix} $$

We can even rewrite this answer:

$$ \begin{split} x &=& x_5 ~\pm~ \dfrac{h(y_2 - y_1)}{d} \\ y &=& y_5 ~\pm~ \dfrac{h(x_2 - x_1)}{d} \end{split} $$

Download

Here is the Matlab script used to check the results from this page:

circle_circle_intersection.m