The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Intersections of Circles and Lines Mathematics

In this explainer, we will learn how to find points of intersection of circles and lines.

Suppose we have a circle and a line in the plane. There are three possibilities.

Either the line cuts through the circle, intersecting it in two points; or the line misses the circle, not intersecting it at all; or the line is tangent to the circle, intersecting it in one point.

Recall that the equation of a straight line may be given in the form 𝑦=π‘šπ‘₯+𝑏, where π‘š is the line’s slope and 𝑏 its 𝑦-intercept, and that the equation of a circle may be given in standard form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘Ÿ, where (β„Ž,π‘˜) is the center of the circle and π‘Ÿ its radius, or in general form π‘₯+𝑦+𝐷π‘₯+𝐸𝑦+𝐹=0, with constants 𝐷, 𝐸, and 𝐹.

If the line and the circle intersect at a point 𝑃, then the coordinates of 𝑃 simultaneously solve the equations of both the line and the circle. This means that we can substitute β€œπ‘šπ‘₯+𝑏” for 𝑦 in the circle equation to calculate the coordinates of the point of intersection 𝑃. The result of this substitution is to eliminate 𝑦 from the circle equation, leaving us with a quadratic equation in π‘₯: 𝐴π‘₯+𝐡π‘₯+𝐢=0.

The roots of this quadratic equation are precisely the π‘₯-coordinates of the intersection points of the line with the circle. The number of roots a quadratic equation has over the real numbers is controlled by its discriminant Ξ”=π΅βˆ’4𝐴𝐢.

Thus, the number of times a given straight line intersects a circle can be read off from the discriminant of the associated quadratic. When Ξ”>0, the line intersects the circle twice; when Ξ”=0, the line intersects the circle once (i.e., it is tangent to the circle); and when Ξ”<0, the line and the circle are disjoint.

Example 1: Determining How Many Times a Line Intersects a Circle

Consider the circle (π‘₯βˆ’5)+(𝑦+2)=8 and the lines πΏβˆΆπ‘¦=βˆ’π‘₯βˆ’1 and πΏβˆΆπ‘¦=π‘₯βˆ’7. How, if at all, do 𝐿 and 𝐿 intersect the circle?

Answer

To work out how a line 𝑦=π‘šπ‘₯+𝑏 intersects a circle (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨, we proceed as follows:

  1. Substitute π‘šπ‘₯+𝑏 into (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨, eliminating 𝑦.
  2. This results in a quadratic equation in π‘₯, 𝐴π‘₯+𝐡π‘₯+𝑐=0. Calculate the discriminant of this quadratic, Ξ”=π΅βˆ’4𝐴𝐢.
  3. If Ξ”>0, then the line intersects the circle in two points. If Ξ”=0, then the line is tangent to the circle. If Ξ”<0, then the line and the circle are disjoint.

We will deal with the two lines one at a time. First, we substitute πΏβˆΆπ‘¦=βˆ’π‘₯βˆ’1 into (π‘₯βˆ’5)+(𝑦+2)=8: (π‘₯βˆ’5)+(𝑦+2)=8(π‘₯βˆ’5)+(βˆ’π‘₯βˆ’1+2)=8π‘₯βˆ’10π‘₯+25+π‘₯βˆ’2π‘₯+1=82π‘₯βˆ’12π‘₯+18=0.

We now compute the discriminant of the resulting quadratic: Ξ”=π΅βˆ’4𝐴𝐢=(βˆ’12)βˆ’4Γ—2Γ—18=144βˆ’144=0, which tells us that the line meets the circle in one point, or, in other words, 𝐿 is tangent to (π‘₯βˆ’5)+(𝑦+2)=8.

Now, for πΏβˆΆπ‘¦=π‘₯βˆ’7, we follow exactly the same procedure. First, we substitute: (π‘₯βˆ’5)+(𝑦+2)=8(π‘₯βˆ’5)+(π‘₯βˆ’7+2)=8π‘₯βˆ’10π‘₯+25+π‘₯βˆ’10π‘₯+25=82π‘₯βˆ’20π‘₯+42=0.

Then, we check the discriminant: Ξ”=π΅βˆ’4𝐴𝐢=(βˆ’20)βˆ’4Γ—2Γ—42=400βˆ’336=64.

The positive discriminant tells us that the quadratic has two real roots and that the line 𝐿 intersects the circle in two distinct points.

So, to answer the question, 𝐿 is tangent to the circle, while 𝐿 intersects it in two points.

Now that we know how to work out if and how many times a line intersects a circle, let us turn our attention to calculating the coordinates of the intersection points.

Given the equation of a circle (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨ and the equation of a line 𝑦=π‘šπ‘₯+𝑏, we proceed as before by substituting π‘šπ‘₯+𝑏 in for 𝑦 (or π‘¦βˆ’π‘π‘š for π‘₯; the result will be the same) in the equation of the circle to obtain a quadratic equation in π‘₯. This time, however, instead of computing the discriminant of the quadratic, we are going to solve it in order to get the π‘₯-coordinates of the intersection points of the line and the circle. We can then substitute these π‘₯-coordinates back into the equation of the line to get the 𝑦-coordinates of the intersection points.

Example 2: Finding the Coordinates of the Points Where a Line Intersects a Circle given in Standard Form

Find the coordinates of the points where the circle (π‘₯+2)+(π‘¦βˆ’4)=5 meets the line 𝑦=βˆ’2π‘₯+4.

Answer

The first step is to substitute 𝑦=βˆ’2π‘₯+4 into (π‘₯+2)+(π‘¦βˆ’4)=5 to obtain a quadratic equation in π‘₯: (π‘₯+2)+(π‘¦βˆ’4)=5(π‘₯+2)+(βˆ’2π‘₯+4βˆ’4)=5π‘₯+4π‘₯+4+(βˆ’2π‘₯)=55π‘₯+4π‘₯βˆ’1=0.

We now solve this quadratic to find the π‘₯-coordinates of the intersection points of the line and the circle. In this example, the quadratic expression 5π‘₯+4π‘₯βˆ’1 factors immediately, as 5π‘₯+4π‘₯βˆ’1=5π‘₯+5π‘₯βˆ’π‘₯βˆ’1=5π‘₯(π‘₯+1)βˆ’(π‘₯+1)=(5π‘₯βˆ’1)(π‘₯+1).

Thus, (5π‘₯βˆ’1)(π‘₯+1)=0 gives us π‘₯-coordinates of π‘₯=15 and π‘₯=βˆ’1. We can now substitute these values back into the linear equation 𝑦=βˆ’2π‘₯+4 to find the corresponding 𝑦-coordinates 𝑦=βˆ’2π‘₯+4𝑦=βˆ’2Γ—15+4𝑦=185 and 𝑦=βˆ’2π‘₯+4𝑦=βˆ’2Γ—(βˆ’1)+4𝑦=6.

Therefore, the coordinates of the intersection points of the circle (π‘₯+2)+(π‘¦βˆ’4)=5 and the line 𝑦=βˆ’2π‘₯+4 are ο€Ό15,185 and (βˆ’1,6).

The final step of the previous example was to substitute the π‘₯-coordinates of the intersection points back into the linear equation 𝑦=βˆ’2π‘₯+4 to find the corresponding 𝑦-coordinates. You might like to think about what happens if we instead substitute into the equation of the circle (π‘₯+2)+(π‘¦βˆ’4)=5 and why.

The previous example dealt with finding the intersection points of a line and a circle given in standard form. Let us now look at an example when the circle is given in general form.

Example 3: Finding the Coordinates of the Points Where a Line Intersects a Circle given in General Form

Find the coordinates of the points where the circle π‘₯+𝑦+2π‘₯βˆ’2π‘¦βˆ’3=0 meets the line 𝑦=12π‘₯+32.

Answer

The method for finding the intersection points of a line 𝑦=π‘₯π‘š+𝑏 and a circle given in general form is as follows:

  1. Substitute π‘šπ‘₯+𝑏 for 𝑦 in the equation of the circle, here π‘₯+𝑦+2π‘₯βˆ’2π‘¦βˆ’3=0.
  2. This results in a quadratic equation in π‘₯, 𝐴π‘₯+𝐡π‘₯+𝑐=0. The solutions to this quadratic equation are the π‘₯-coordinates of the points where the line and the circle intersect.
  3. Having calculated the π‘₯-coordinates, we substitute these values back into the equation of the line to find the corresponding 𝑦-coordinates.

So, we begin by substituting 𝑦=12π‘₯+32 into the circle equation π‘₯+𝑦+2π‘₯βˆ’2π‘¦βˆ’3=0: π‘₯+𝑦+2π‘₯βˆ’2π‘¦βˆ’3=0π‘₯+ο€Ό12π‘₯+32+2π‘₯βˆ’2ο€Ό12π‘₯+32οˆβˆ’3=0.

Now, we expand and simplify: π‘₯+ο€Ό12π‘₯+32+2π‘₯βˆ’2ο€Ό12π‘₯+32οˆβˆ’3=0π‘₯+14π‘₯+2Γ—32Γ—12π‘₯+ο€Ό32+2π‘₯βˆ’2Γ—12π‘₯βˆ’2Γ—32βˆ’3=054π‘₯+52π‘₯βˆ’154=054ο€Ήπ‘₯+2π‘₯βˆ’3=0.

We can solve the resulting quadratic equation to find the π‘₯-coordinates of the intersection points of the line and the circle. In this example, the quadratic 54ο€Ήπ‘₯+2π‘₯βˆ’3=0 factorizes as 54(π‘₯βˆ’1)(π‘₯+3)=0, giving us the π‘₯-coordinates of π‘₯=1 and π‘₯=βˆ’3. Finally, we substitute these values back into the linear equation 𝑦=12π‘₯+32 to find the 𝑦-coordinates of the intersection points: 𝑦=12π‘₯+32𝑦=12Γ—1+32𝑦=2 and 𝑦=12π‘₯+32𝑦=12Γ—(βˆ’3)+32𝑦=0.

Therefore, the coordinates of the points of intersection of the circle π‘₯+𝑦+2π‘₯βˆ’2π‘¦βˆ’3=0 and the line 𝑦=12π‘₯+32 are (1,2) and (βˆ’3,0).

We will now use what we have learned about intersections of lines and circles to solve problems involving unknown constants in circle and line equations.

Example 4: Finding the Possible Unknowns in the Equation of a Circle given a Line that Touches it at a Given Point

The line 𝑦+2π‘₯=5 is tangent to the circle (π‘₯βˆ’π‘)+(π‘¦βˆ’2)=5, where 𝑝 is a constant. Work out the two possible values of 𝑝.

Answer

Since we are given that the line 𝑦+2π‘₯=5 is tangent to the circle (π‘₯βˆ’π‘)+(π‘¦βˆ’2)=5, we know that there exists a point at which the line and the circle meet; in particular, at this point, the value of 𝑦 in the equation of the line will be equal to the value of 𝑦 in the equation of the circle. That is to say, it is valid to substitute an expression for 𝑦 derived from the equation of the line into the equation of the circle. In order to do this, we first make 𝑦 the subject of 𝑦+2π‘₯=5: 𝑦+2π‘₯=5𝑦=βˆ’2π‘₯+5.

Now, we can substitute and simplify: (π‘₯βˆ’π‘)+(π‘¦βˆ’2)=5(π‘₯βˆ’π‘)+(βˆ’2π‘₯+5βˆ’2)=5π‘₯βˆ’2𝑝π‘₯+𝑝+4π‘₯βˆ’12π‘₯+9=55π‘₯βˆ’(12+2𝑝)π‘₯+𝑝+4=0.

The roots of the resulting quadratic 5π‘₯βˆ’(12+2𝑝)π‘₯+𝑝+4=0 are the π‘₯-coordinates of the intersection points of the line 𝑦+2π‘₯=5 and the circle (π‘₯βˆ’π‘)+(π‘¦βˆ’2)=5. However, we are told that this line is tangent to the circle; that is to say, it meets the circle in exactly one point. Therefore, the quadratic 5π‘₯βˆ’(12+2𝑝)π‘₯+𝑝+4=0 must have exactly one real root. This tells us that the discriminant Ξ”=π΅βˆ’4𝐴𝐢 must be zero. Let us write this down: Ξ”=π΅βˆ’4𝐴𝐢=0(βˆ’(12+2𝑝))βˆ’4Γ—5×𝑝+4=0144+48𝑝+4π‘βˆ’20π‘βˆ’80=0βˆ’16𝑝+48𝑝+64=0βˆ’16ο€Ήπ‘βˆ’3π‘βˆ’4=0.

The roots of this quadratic βˆ’16ο€Ήπ‘βˆ’3π‘βˆ’4=0 will give us the values of 𝑝 that we are looking for. Factorizing, we have βˆ’16ο€Ήπ‘βˆ’3π‘βˆ’4=βˆ’16(𝑝+1)(π‘βˆ’4)=0, and so 𝑝=βˆ’1 or 𝑝=4. These two values are the π‘₯-coordinates of the centers of the two circles of radius √5 whose centers lie on the line 𝑦=2 and that are tangent to the line 𝑦+2π‘₯=5.

We will finish with another problem-solving example involving circles, lines, and their intersections.

Example 5: Finding the Range of Unknowns That Lead to a Line Not Intersecting a Circle

Consider the circle (π‘₯βˆ’5)+(π‘¦βˆ’3)=5 and the line 𝑦=π‘Žπ‘₯+3, where π‘Ž is a constant. Find the range of values of π‘Ž for which the circle and the line are disjoint.

Answer

If we substitute 𝑦=π‘Žπ‘₯+3 into the equation of the circle (π‘₯βˆ’5)+(π‘¦βˆ’3)=5, we will get a quadratic equation whose solutions are the π‘₯-coordinates of the points where the line and the circle intersect. Since we are trying to find a range of values of π‘Ž for which the line and the circle do not intersect, we are aiming for a quadratic equation with no real solutions.

So, we start by substituting: (π‘₯βˆ’5)+(π‘¦βˆ’3)=5(π‘₯βˆ’5)+(π‘Žπ‘₯+3βˆ’3)=5.

Expanding and simplifying, π‘₯βˆ’10π‘₯+25+π‘Žπ‘₯βˆ’5=0ο€Ήπ‘Ž+1π‘₯βˆ’10π‘₯+20=0, we obtain the quadratic ο€Ήπ‘Ž+1π‘₯βˆ’10π‘₯+20=0. We are interested in the values of π‘Ž for which the discriminant Ξ”=π΅βˆ’4𝐴𝐢 of this quadratic is negative. That is, (βˆ’10)βˆ’4Γ—ο€Ήπ‘Ž+1×20<0100βˆ’80π‘Žβˆ’80<0βˆ’80π‘Ž<βˆ’2080π‘Ž>20π‘Ž>14.

The ranges of values for which π‘Ž>14 are π‘Ž>12 when π‘Ž is positive and π‘Ž<βˆ’12 when π‘Ž is negative.

These ranges of values give ranges of slopes of the line 𝑦=π‘Žπ‘₯+3 with 𝑦-intercept 3 for which it does not intersect the circle (π‘₯βˆ’5)+(π‘¦βˆ’3)=5.

The pink region contains all the lines with the negative slope π‘Ž<βˆ’12, while the blue region contains all the lines with positive slope π‘Ž>12.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • A line and a circle in the plane can interact in one of three ways: they may intersect in two points, they may intersect in one point, or they may not intersect at all. If the line and the circle do intersect, then the coordinates of their point (or points) of intersection simultaneously solve the equation of the line and the equation of the circle. This allows us to use algebra to calculate whether a line and a circle intersect and, if they do, the coordinates of their points of intersection.
  • Given the equation of a line 𝑦=π‘šπ‘₯+𝑏 and the equation of a circle in either standard form (π‘₯βˆ’β„Ž)+(π‘¦βˆ’π‘˜)=π‘ŸοŠ¨οŠ¨οŠ¨ or general form π‘₯+𝑦+𝐷π‘₯+𝐸𝑦+𝐹=0, we can substitute the expression π‘šπ‘₯+𝑏 for 𝑦 in the equation of the circle to obtain a quadratic equation in π‘₯, 𝐴π‘₯+𝐡𝑋+𝐢=0.
  • The discriminant Ξ”=π΅βˆ’4𝐴𝐢 of the quadratic 𝐴π‘₯+𝐡𝑋+𝐢=0 tells us about the intersections of the line and the circle. If Ξ”>0, then the line and the circle intersect in two points. If Ξ”=0, then the line is tangent to the circle. If Ξ”<0, then the line and the circle are disjoint.
  • We can solve the quadratic 𝐴π‘₯+𝐡𝑋+𝐢=0 to find the π‘₯-coordinates of the points of intersection of the line and the circle. We can then substitute these values back into the equation of the line 𝑦=π‘šπ‘₯+𝑏 to find the 𝑦-coordinates of the points of intersection.
  • We can use conditions of the discriminant (i.e., whether it is positive, negative, or zero) to find unknown constants in equations of lines and circles when we know how they intersect.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.