Intersection of a line and a circle

Introduction

In this article it will be explained how to calculate the coordinates of the points ($P=\left(x_p,y_p\right)$) where a line and a circle intersect. The required equations will be derived and an interactive application is provided to calculate the intersection points of a given line and circle. A line and a circle have no intersection points, one intersection point (line is tangent to circle) or two intersection points 🙄.

JavaScript application

Line equation

Let a, b and c be the line parameters in the general linear equation:

$$a\; <!--\; \; -->x+b\; <!--\; \; -->y+c=0$$

We can rewrite this general form of the equation to the so called slope-intercept form, which is generally the form for representing an equation of a straight line in the Euclidean plane:

$$\begin{array}{llll}\text{If}& b\ne \; <!--\; \ne \; -->0& \text{:}& y=m\; <!--\; \; -->x+y_0\\ \text{If}& b=0& \text{:}& x=k\end{array}$$ with:
$$\begin{array}{lll}m=-\; <!--\; -\; -->\frac{a}{b}\text{,}& y_0=-\; <!--\; -\; -->\frac{c}{b}\text{,}& k=-\; <!--\; -\; -->\frac{c}{a}\end{array}$$ [1]

Here m is called the slope or gradient of the line and y₀ is called the y-intercept. If b = 0, the line is a vertical line ($x=k$).

Circle equation

Let $M=\left(x_m,y_m\right)$ be the circle center and r be the circle radius in the general circle equation:

$$\begin{array}{c}{\left(x-\; <!--\; -\; -->x_m\right)}^2+{\left(y-\; <!--\; -\; -->y_m\right)}^2=r^2\\ \iff \\ x^2-\; <!--\; -\; -->2\; <!--\; \; -->x\; <!--\; \; -->x_m+{x_m}^2+y^2-\; <!--\; -\; -->2\; <!--\; \; -->y\; <!--\; \; -->y_m+{y_m}^2-\; <!--\; -\; -->r^2=0\end{array}$$ [2]

Intersection points

Given that $y=-\; <!--\; -\; -->\frac{a}{b}\; <!--\; \; -->x-\; <!--\; -\; -->\frac{c}{b}$ (equation [1]), means that the coordinates of the intersection points of a non-vertical line and a circle are $P=\left(x_p,-\; <!--\; -\; -->\frac{a}{b}\; <!--\; \; -->x_p-\; <!--\; -\; -->\frac{c}{b}\right)$ (if any exists).

When the line is vertical ($x=-\; <!--\; -\; -->\frac{c}{a}$), the coordinates of the intersection points of a line and a circle are $P=\left(-\; <!--\; -\; -->\frac{c}{a},y_p\right)$ (if any exists).

Intersection with non-vertical line

A point where a line and circle intersect (or touch) must be a point belonging to both the line and the circle. Substituting line equation [1] ($y=m\; <!--\; \; -->x+y_0$) in circle equation [2] yields an equation in which x represents the x-coordinates of the intersection points (if any exists). Substitution results in an equation of the form:

$$A\; <!--\; \; -->x^2+B\; <!--\; \; -->x+C=0$$

Solving this equation for x gives us the x-coordinates x_p:

$$x_p=\frac{-\; <!--\; -\; -->B\pm \; <!--\; \pm \; -->\sqrt{B^2-\; <!--\; -\; -->4\; <!--\; \; -->A\; <!--\; \; -->C}}{2\; <!--\; \; -->A}$$ with: $$A=\frac{a^2}{b^2}+1$$ $$B=2\; <!--\; \; -->\left(\frac{ac}{b^2}-\; <!--\; -\; -->x_m+\frac{a}{b}\; <!--\; \; -->y_m\right)$$ $$C={x_m}^2+{y_m}^2+\frac{2\; <!--\; \; -->c}{b}\; <!--\; \; -->y_m+\frac{c^2}{b^2}-\; <!--\; -\; -->r^2$$

Let $D=B^2-\; <!--\; -\; -->4\; <!--\; \; -->A\; <!--\; \; -->C$ :

• If D < 0, then x_p has no solutions; line and circle do not intersect.
• If D = 0, then x_p has exactly one solution; line is tangent to circle.
• If D > 0, then x_p has two solutions; line and circle intersect at two points.

Intersection with vertical line

Substituting vertical line equation x = k in circle equation [2] results in an equation of the form:

$$y^2+B\; <!--\; \; -->y+C=0$$

Solving this equation for y gives us the y-coordinates y_p:

$$y_p=\frac{-\; <!--\; -\; -->B\pm \; <!--\; \pm \; -->\sqrt{B^2-\; <!--\; -\; -->4\; <!--\; \; -->C}}{2}$$ with: $$B=-\; <!--\; -\; -->2\; <!--\; \; -->y_m$$ $$C={x_m}^2+{y_m}^2+\frac{2\; <!--\; \; -->c}{a}\; <!--\; \; -->x_m+\frac{c^2}{a^2}-\; <!--\; -\; -->r^2$$

Let $D=B^2-\; <!--\; -\; -->4\; <!--\; \; -->C$ :

• If D < 0, then y_p has no solutions; line and circle do not intersect.
• If D = 0, then y_p has exactly one solution; line is tangent to circle.
• If D > 0, then y_p has two solutions; line and circle intersect at two points.

Task

Calculate the intersection points (if they exist) of $y=3\; <!--\; \; -->x-\; <!--\; -\; -->2$ and ${\left(x-\; <!--\; -\; -->6\right)}^2+y^2=36$.