Intersection of a circumference and a straight line

We are going to study the relative positions where a straight line and a circumference can be in the same plane.

To do so, we will name several points, straight lines and segments that are singular in the circumference:

• Center is an interior point equidistant from all the points of the circumference.
• Radius is the distance from the center to any point of the circumference.
• Chord is the segment that joins two points of the circumference; the chords of maximum length are the diameters.
• Secant straight line is any line that cuts the circumference in two points.
• Tangent straight line is any line that touches the circumference at only one point.
• Touching point is the point where the tangent touches the circumference.

To find the common points of a circumference and a straight line we will solve the system formed by their equations. Namely if we have:

• the circumference given by the equation $(x-a{)}^2+(y-b{)}^2=r^2$ or by the equation $x^2+y^2+Ax+By+C=0$
• the straight line given by the general equation of a straight line: $y-y_0=m\cdot (x-x_0)$

What we must solve is one of two following systems (depending on the circumference given to us): $$\{\begin{array}{l}(x-a{)}^2+(y-b{)}^2=r^2\\ y-y_0=m\cdot (x-x_0)\end{array}\text{ or }\{\begin{array}{l}{x}^2+y^2+Ax+By+C=0\\ y-y_0=m\cdot (x-x_0)\end{array}$$

Note that once we know one of the two equations that define a circumference we can obtain the other one. Thus, we can consider the case: $$\{\begin{array}{l}{x}^2+y^2+Ax+By+C=0\\ y-y_0=m\cdot (x-x_0)\end{array}$$ Isolating $y$ from the straight line we obtain: $$y=y_0+m\cdot (x-x_0)$$ and replacing this expression in the general equation of the circumference we obtain: $$x^2+(y_0+m\cdot (x-x_0){)}^2+Ax+B(y_0+m\cdot (x-x_0))+C=0$$ Manipulating the equation we will obtain: $$\begin{array}{l}{x}^2+(y_0+m\cdot (x-x_0){)}^2+Ax+B(y_0+m\cdot (x-x_0))+C=0\\ x^2+y_0^2+2\cdot y_0\cdot m\cdot x-2\cdot y_0\cdot m\cdot x_0+m^2\cdot (x-x_0{)}^2+\\ +Ax+B{y}_0+B\cdot m\cdot x-B\cdot m\cdot x_0+C=0\\ x^2+m^2\cdot x^2+2\cdot y_0\cdot m\cdot x-2\cdot m^2\cdot x\cdot x_0+Ax+B\cdot m\cdot x+\\ +y_0^2-2\cdot y_0\cdot m\cdot x_0+B\cdot y_0-B\cdot m\cdot x_0+m^2\cdot x_0^2+C=0\\ x^2(1+m^2)+x(2\cdot y_0\cdot m-2\cdot m^2\cdot x_0+A+B\cdot m)+\\ +y_0^2-2\cdot y_0\cdot m\cdot x_0+B\cdot m\cdot x_0+m^2\cdot x_0^2+C=0\end{array}$$

which is an equation of the second grade in the variable $x$.

Since we have a general equation of the second degree, we know that depending on the sign of the discriminant ($\mathrm{\Delta }=b^2-4ac$), we will have the following solutions:

• If $\mathrm{\Delta }>0$ we have two solutions: then the straight line and the circumference are secant.
• If $\mathrm{\Delta }=0$ we only have one solution: then the straight line and the circumference are tangent.
• If $\mathrm{\Delta }<0$ we have no solution: then the straight line and the circumference are exterior. Therefore they do not meet at any point.

The different possibilities can be seen in the following picture: