Intersection of a line and a circle

Introduction

Given: line parameters a, b and c and circle center $M=\left(x_m,y_m\right)$ and circle radius r.

General linear equation with two variables:

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

We can write this as an equation of a straight line like:

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

In case b ≠ 0, 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.

General circle equation:

$${\left(x-<!--\; -\; -->x_m\right)}^2+{\left(y-<!--\; -\; -->y_m\right)}^2=r^2$$ ⇔ $$x^2-<!--\; -\; -->2\; <!--\; \; -->x\; <!--\; \; -->x_m+{x_m}^2+y^2-<!--\; -\; -->2\; <!--\; \; -->y\; <!--\; \; -->y_m+{y_m}^2+-<!--\; -\; -->r^2=0$$ [1]

If b ≠ 0:

Substituting line equation $y=m\; <!--\; \; -->x+y_0$ in [1] results in an equation of the form:

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

This equation's solutions for x are the x-coordinates of the intersection points $P=\left(x_p,-<!--\; -\; -->\frac{a}{b}\; <!--\; \; -->x_p-<!--\; -\; -->\frac{c}{b}\right)$ :

$$x_p=\frac{-<!--\; -\; -->B\pm <!--\; \pm \; -->\sqrt{B^2-<!--\; -\; -->4\; <!--\; \; -->A\; <!--\; \; -->C}}{2\; <!--\; \; -->A}$$ 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 does 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 intersects at two points.

If b = 0:

Substituting vertical line equation x = k in [1] results in an equation of the form:

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

This equation's solutions for y are the y-coordinates of the intersection points $P=\left(-<!--\; -\; -->\frac{c}{a},y_p\right)$ :

$$y_p=\frac{-<!--\; -\; -->B\pm <!--\; \pm \; -->\sqrt{B^2-<!--\; -\; -->4\; <!--\; \; -->C}}{2}$$ 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 does 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 intersects at two points.