Intersection of Line and Circle: Explanation & Examples

A line is a straight (1)-dimensional figure consisting of a set of points extending infinitely to either side. A circle is a simple closed figure made of a set of points. All the points on its boundary are at a fixed distance from the circle’s centre.
A point at which two or more objects meet or contact one other is an intersection in mathematics.

This article will discuss three ways of intersection between a line and a circle and two different methods to find the point of intersection of a line and circle. We will also have some solved examples on this topic.

General Form of Circle and Line

Before diving into the topics, let us learn some general details about a line and a circle.
The general form of a line can be given by $ax+by+c=0.$

The general equation of the circle with centre $\left(h,k\right)$ and radius $r$ is ${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2$

The distance from a point $\left(m,n\right)$ to the line, $ax+by+c=0$ is given by $d=\frac{\left|am+bn+c\right|}{\sqrt{a^2+b^2}}$

Intersection Between Circle and Line

A line can intersect a circle in three possible ways, as shown below:
1. We obtain two points of the intersection if a line intersects or cuts through the circle, as shown in the diagram below.

We can see that in the above figure, the line meets the circle at two points. This line is called the secant to the circle.
2. If we draw a tangent line to the circle, we will only have one point of intersection, as shown in the diagram:

3. There will be no point of intersection if a line does not touch the circle at all.

In the above figure, the line does not even come close to touching the circle. As a result, we may conclude that no point of intersection exists.

Methods to Find the Point of Intersection of a Line and Circle

There are two methods to think about this.
Method 1: Let us consider the equation of the circle be $x^2+y^2=a^2.$ And that of the line be $y=mx+c.$
First, if we want to solve the two equations in two unknowns, we need to frame a quadratic equation in $x.$
Substitute the linear equation in the circle’s equation. Linear equations are often defined in terms of $y.$ We’ll replace the $y$-values of the circle equation with the $y$ value of the linear equation. So, we get
$x^2+{\left(mx+c\right)}^2=a^2$
Now, simplify the equation to get a quadratic equation. And factorise the quadratic equation by using the algebraic identity ${\left(a+b\right)}^2=a^2+2ab+b^2$
$x^2+{\left(mx\right)}^2+2mxc+c^2=a^2$
$x^2+m^2{x}^2+2mcx+c^2=a^2$
$\left(1+m^2\right)x^2+2mcx+c^2–a^2=0$
$x=\frac{–{\left(2mc\right)}^2\pm \sqrt{{\left(2mc\right)}^2–4\times \left(1+m^2\right)\times \left(c^2–a^2\right)}}{2\left(1+m^2\right)}$
Or $x=\frac{–{\left(2mc\right)}^2\pm \sqrt{D}}{2\left(1+m^2\right)}$
Where $D={\left(2mc\right)}^2–4\times \left(1+m^2\right)\times \left(c^2–a^2\right)$
Find roots, i.e., the values of $x.$ Substitute the values of $x$ in the linear equation to get the corresponding values of $y.$ In terms of geometry, what do the roots of this quadratic mean?
Now, if you recall, when we were talking about equations in general, the $x$ and the $y$ in equations represent the $x$ and the $y$ coordinates of all points on the curve. Therefore, the roots of the equation will represent the $x$ coordinates of the points of intersections of the line and the circle.
In fact, by solving any two equations in coordinate geometry, we obtain the $x$ (or $y$) coordinates of the points of intersections of the two curves represented by the equations.
The line will meet the circle at two distinct points if we obtain two distinct real roots. If we have two coincident roots, we know that the line only touches the circle at one point (i.e. two coincident points). Finally, if the formed equation has no actual roots, the line will not touch or cross the circle.
And how are you going to figure it out?
The discriminant of a quadratic equation determines the nature of its roots. So all we have to do now is determine the quadratic equation’s discriminant and check its sign.
A positive sign indicates that the line intersects or touches the circle at two different locations, a zero sign indicates tangency, and a negative sign indicates that the line does not intersect or touch the circle.
1. If $D>0,$ then there are two points of intersection
2. If $D=0,$ then there is one point of intersection.
3. If $D<0,$ then there is no point of intersection.
To explain everything I just said, here’s a diagram.

In the case of other conic sections, such as parabola, ellipse, and hyperbola, we may apply the same approach to determine the location of a line.
Method 2: To determine the position of a line with respect to a circle, we’ll find its distance from the centre of the circle. Let $d$ be this distance and $r$ be the radius of the circle. Then,
1. If the distance is less than the radius, i.e., $d<r,$ the line must intersect the circle at two distinct points.

2. If the distance equals the radius, i.e., $d=r,$ the line will touch the circle at only one point.

Learn Concepts on Circles

3. If the distance is greater than the radius, i.e., $d>r,$ the line will lie entirely outside the circle.

Solved Examples: Intersection Between Circle and Line

Q.1. Prove that the line $y=x+4$ intersects the circle $x^2+y^2+8x+2y–84=0.$
Ans: We are given a linear equation $y=x+4.$
The equation of a circle is $x^2+y^2+8x+2y–84=0.$
Substitute $y=x+4$ in the equation of the circle $x^2+y^2+8x+2y–84=0.$
$x^2+{\left(x+4\right)}^2+8x+2\left(x+4\right)–84=0$
We know that ${\left(x+4\right)}^2=x^2+8x+16$
$x^2+x^2+8x+16+8x+2\left(x+4\right)–84=0$
Combine the like terms $2x^2+18x+24–84=0$
Divide the entire equation by a constant $2$ to simplify it:
$x^2+9x–30=0$
$x=\frac{–9\pm \sqrt{81+120}}{2}=\frac{–9\pm 14.17}{2}$
$x=\frac{–9+14.17}{2}=\frac{5.17}{2}=2.585$
or $x=\frac{–9–14.17}{2}=\frac{–23.17}{2}=–11.585$
In this case, the roots of the equation are real and distinct. Hence the line crosses the circle at two distinct points.

Q.2. Prove that the line $y=3x+1$ intersects the circle $x^2+y^2+3x+4y–18=0.$
Ans: We are given a linear equation $y=3x+1.$
The equation of a circle is $x^2+y^2+3x+4y–18=0.$
Substitute $y=3x+1$ in the equation of the circle $x^2+y^2+3x+4y–18=0.$
$x^2+{\left(3x+1\right)}^2+3x+4\left(3x+1\right)–18=0$
We know that ${\left(3x+1\right)}^2=9x^2+6x+1$
$x^2+9x^2+6x+1+3x+4\left(3x+1\right)–18=0$
Combine the like terms $10x^2+21x–13=0$
simplify the equation: $x=\frac{–21\pm \sqrt{{21}^2–4\times 10\times \left(–13\right)}}{20}=\frac{–21\pm \sqrt{441+520}}{20}=\frac{–21\pm \sqrt{961}}{20}$
$x=\frac{–21\pm 13}{20}$
$x=\frac{–21+13}{20}=\frac{–2}{5}$ and $x=\frac{–21–13}{20}=\frac{–17}{10}$
In this case, the roots of the equation are real and distinct. Hence the line crosses the circle at two distinct points.

Q.3. Determine whether the given line intersects the given circle at two distinct points, touch the circle or does not intersect the circle at any point: $L:3x+4y=10;C:x^2+y^2=9$
Ans: We are given a linear equation $3x+4y=10$
The equation of a circle is $x^2+y^2=9$
Substitute $y=\frac{10–3x}{4}$ in the equation of the circle $x^2+y^2=9$
$x^2+{\left(\frac{10–3x}{4}\right)}^2=9$
$16x^2+100–60x+9x^2=144$
Combine the like terms $25x^2–60x–44=0$
The discriminant of the equation equals $D=3600+4400=8000,$ which is positive.
Hence, the line will intersect the circle at two distinct points.

Q.4. Determine whether the given line intersects the given circle at two distinct points, touch the circle or does not intersect the circle at any point:$L:x=3;C:x^2+y^2+4x+6y–3=0$
Ans: We are given a linear equation $x=3.$
The equation of a circle is $x^2+y^2+4x+6y–3=0$
We know that the general equation of the circle with centre $\left(h,k\right)$ and radius $r$ is
${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2$
Converting the given equation in the standard form, we get
$x^2+4x+4–4+y^2+6y+9–9–3=0$
${\left(x+2\right)}^2+{\left(y+3\right)}^2=16$
Therefore, the centre of the circle is $\left(–2,–3\right)$ and the radius of the circle is $4.$
The distance from a point $\left(m,n\right)$ to the line, $Ax+By+C=0$ is given by: $d=\frac{\left|Am+Bn+C\right|}{\sqrt{A^2+B^2}}$
We’ll calculate the distance between the given line $\left(x=3\right)$ and the centre $\left(–2,–3\right),$ which is $d=\frac{\left|–2+0–3\right|}{1}=5$
And $d>r,$ so the line will not cross or touch the circle.

Q.5. Consider the equation of the circle $x^2+y^2–2x+4y=0.$ Find the value(s) of $‘c^{\prime }$ for which the line $x+2y=c$
(i) touches the circle
(ii) crosses the circle at two distinct points
(iii) does not cross the circle at any point
Ans: Given, equation of the circle $x^2+y^2–2x+4y=0.$
We know that the general equation of the circle with centre $\left(h,k\right)$ and radius $r$ is
${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2$
Converting the given equation in the standard form, we get
$x^2–2x+1–1+y^2+4y+4–4=0$
${\left(x–1\right)}^2+{\left(y+2\right)}^2=5$
Therefore, the centre of the circle is $\left(1,–2\right)$ and radius of the circle is $\sqrt{5}.$
We must determine the values of a parameter, $‘c^{\prime }.$
The distance from a point $\left(m,n\right)$ to the line, $Ax+By+C=0$ is given by:$d=\frac{\left|Am+Bn+C\right|}{\sqrt{A^2+B^2}}$
To begin, let us calculate the distance of the given line $x+2y=c$ from the centre $C\left(1,–2\right):$
$CP=\frac{\left|1+2\times \left(–2\right)+c\right|}{\sqrt{1^2+2^2}}=\frac{\left|c+3\right|}{\sqrt{5}}$
(i) In this case, $CP=r,$
or $\frac{\left|c+3\right|}{\sqrt{5}}=\sqrt{5}\Rightarrow \left|c+3\right|=5$
Therefore $c=2$ or $-8.$
(ii) Here, $CP<r,$
or $\frac{\left|c+3\right|}{\sqrt{5}}<\sqrt{5}\Rightarrow \left|c+3\right|<5.$
Therefore, $–8<c<2.$
(iii) In this case, $CP>r,$
or $\frac{\left|c+3\right|}{\sqrt{5}}>\sqrt{5}\Rightarrow \left|c+3\right|>5.$
Therefore, the values of $c$ will be $c<–8$ or $c>2.$

Summary

In this article, we have discussed line and circle and their general forms. Then we saw the three cases of the intersection of a circle and a line. Also, we discussed the two methods of finding the intersection of a circle and a line in detail.

FAQs on Circle Line Intersection

Q.1. What does it mean for a line to intersect a circle at one point?
Ans: If a line intersects a circle at only one point, that line will be a tangent to the circle.

Q.2. How do you find the intersection of a circle and a line?
Ans: We can find the distance of the line from the centre of the circle. If the distance is less than the radius, the line intersects the circle at two distinct points. If the distance is equal to the radius, then the line touches the circle at one point. If the distance is greater than the radius, then the line never touches the circle.

Q.3. What do you call a line that intersects a circle curve at exactly two points?
Ans: Secant is a line that intersects a circle at exactly two points.

Q.4. What is the standard equation of a circle?
Ans: The standard equation of the circle with centre $\left(h,k\right)$ and radius $r$ is ${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2.$

Q.5. What is the formula to find the distance of a point and a straight line?
Ans: The distance from a point $\left(m,n\right)$ to the line, $Ax+By+C=0$ is given by: $d=\frac{\left|Am+By+C\right|}{\sqrt{A^2+B^2}}.$