In mathematics, an intersection of two or more objects means a point at which these objects intersect or touch each other. A line segment is a straight line with two fixed endpoints, whereas a circle is a figure in which all points are located at a fixed distance from the center of the circle. In this article, we will discuss the intersection of a line and a circle. So, let us get started.

Ways in Which a Line can Intersect a Circle

A line can intersect a circle in the following three ways:

  1. If a line intersects or cuts through the circle, then we will get two points of intersection as shown in figure 1.1. below.
line intersects a circle
Figure 1.1

You can see that in the above figure, the red line intersects the circle at two points. In this case, \triangle > 0.

 

2. If a tangent line is drawn to the circle, then we will have only one point of intersection as shown in the figure below:

Tangent line intersects a circle
Figure 1.2

You can see that the red line is tangent to the green circle as it cuts the circle at a single point only. In this case, \triangle = 0.

 

3. If a line does not touch the circle at all, then there will be no point of intersection as shown in the figure below"

Circles and lines
Figure 1.3

You can see that the red line does not touch the circle at all. Hence, we can say that there is no point of intersection. In this case, \triangle < 0.

 

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Steps for Finding a Point of Intersection of a Line and Circle

If we are given a linear equation of a line and a general equation of a circle, we can easily find whether the line intersects the given circle or not. To do so, we should follow the following steps:

  • Substitute the linear equation into the equation of a circle. Often linear equations are given in terms of y. We will substitute y-values of the equation of a circle with y value of the linear equation.
  • Simplify the equation to get a quadratic equation.
  • Factorize the quadratic equation by expanding the middle term.
  • Find roots, i.e the values of x. Substitute the values of x in the linear equation to get the corresponding y- values.

Note: You can also employ quadratic formula to calculate the roots of the equation.

Now, we will proceed to solve some of the examples of circle-line intersection which will make the above steps clearer.

 

Example 1

Show that the line y = x + 4 intersects the circle x^2 + y^2 + 8x + 2y - 84 = 0, and also determine the point of intersection.

Solution

We are given a linear equation y = x + 4.

The equation of a circle = 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 + (x + 4)^2 + y^2 + 8x + 2(x + 4)^2 - 84 = 0

We know that (x + 4)^2 = x^2 + 8x + 16:

x^2 + x^2 + 8x + 16 + 8x + 2 (x^2 + 8x + 16) - 84= 0

x^2 + x^2 + 8x+ 16 + 8x + 2x^2 + 16x + 32 - 84 = 0

Combine the like terms together:

x^2 + x^2 + 2x^2 + 8x + 8x + 16x +16 + 32 - 84 = 0

4x^2 + 32x - 36 = 0

Divide the entire equation by a constant 4 to simplify it:

x^2 + 8x - 9 = 0

Factorize the above quadratic equation by expanding the middle term:

x^2 + 9x - x - 9 = 0

x (x + 9) - 1 (x + 9 ) = 0

(x - 1) (x + 9) = 0

x = 1 or x = -9

Substitute the values of x, that are  1 and - 9 in the linear equation y = x + 4 to get the value of y.

y = 1 + 4 = 5

y = -9 + 4 = -5

This shows that the line y = x + 4 intersects the circle x^2 + y^2 + 8x + 2y - 84 = 0 at the points (1, 5 ) and ( -9 , -5).

 

Example 2

Show that the line y = 3x + 1 intersects the circle x^2 + y^2 + 3x + 4y - 18 = 0, and also determine the point of intersection.

Solution

We are given a linear equation y = 3x + 1.

The equation of a circle = 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 + (3x + 1)^2  + 3x + 4(3x + 1)^2 - 18 = 0

We know that (3x + 1)^2 = 9x^2 + 6x + 1:

x^2 + 9x^2 + 6x + 1 + 3x + 4(9x^2 + 6x + 1) - 18 = 0

x^2 + 9x^2 + 6x + 1 + 3x + 36x^2 + 24x + 4 - 18 = 0

Combine the like terms together:

x^2 + 9x^2 + 36x^2 + 6x  + 3x  + 24x + 4 + 1 - 18 = 0

46x^2  +33x - 13 = 0

Factorize the above quadratic equation by expanding the middle term:

46x^2 + 46x  -13x - 13 = 0

46x (x + 1) - 13 (x + 1) = 0

(x + 1) (46x - 13) = 0

x = - 1 or x = \frac {13} {46}

Substitute the values of x, that are -1 and \frac {13} {46} in the linear equation y = 3x + 1 to get the value of y.

y = 3 (-1) + 1 = -2

y = 3 (\frac {13} {46}) + 1 = \frac {39} {46} + 1 = \frac {85} {46}

Hence, the line y = 3x+ 1 intersects the circle x^2 + y^2 + 3x + 4y - 18 = 0 at the points (-1, -2) and ( \frac {13}{46} , \frac {85} {46}).

 

Example 3

Determine the points of intersection of the line y = x - 1 and the circle x^2 + y^2 + 4x - 5 = 0.

Solution

We are given a linear equation y = x - 1.

The equation of a circle:

x^2 + y^2 + 4x - 5 = 0.

Substitute y = x - 1 in the equation of the circle x^2 + y^2 + 4x - 5= 0.

x^2 + y^2 + 4x - 5 = 0

x^2 + (x - 1) ^2 + 4x - 5 = 0

x^2 + x^2 - 2x + 1 + 4x - 5 = 0

2x^2  + 2x - 4 = 0

Divide the equation by 2 to simplify it:

x^2 + x - 2 = 0

Factorize the above equation to get the roots of the equation:

x^2 + 2x - x - 2 = 0

x (x + 2) - 1(x + 2) = 0

(x - 1) (x + 2) = 0

x = 1 or x = -2

Substitute the values of x in the equation y = x - 1 to get the y-coordinates of the line:

y = 1 - 1 = 0

y = -2 - 1 = -3

Hence, the line y = x - 1 intersects the circle x^2 + y^2 + 4x - 5 = 0 at points (1, 0) and (- 2, -3).

 

Example 4

Show that the line y = 2x + 1 intersects the circle x^2 + y^2 + 5x + 5y + 2= 0, and also determine the point of intersection.

Solution

We are given a linear equation y = 2x + 1.

The equation of a circle:

x^2 + y^2 + 5x + 5y + 2= 0

Substitute y = 2x + 1 in the equation of the circle x^2 + y^2 + 5x + 5y  + 2= 0.

x^2 + y^2 + 5x  + 5y + 2 = 0

x^2 + (2x + 1) ^2 + 5x + 5 (2x + 1) + 2 = 0

x^2 + 4x^2 + 4x + 1 + 10x + 5 + 2= 0

5x^2  + 14x + 8 = 0

Factorize the above equation to get the roots of the equation:

5x^2 + 10x + 4x + 8 = 0

5x (x + 2) + 4(x + 2) = 0

(5x + 4) (x + 2) = 0

x = -\frac {4}{5} or x = -2

Substitute the values of x in the equation y = 2x + 1 to get the y-coordinates of the line:

y = 2(-2) + 1 = -3

y = 2 (-\frac{4}{5}) - 1 = - \frac {3}{5}

Hence, the line y = 2x + 1 intersects the circle x^2 + y^2 + 5x + 5y  - 1= 0 at points (-2, -3) and (- \frac{4}{5}, -\frac {3}{5}).

 

Example 5

Show that the line y = x + 5 is tangent to the circle x^2 + y^2 + 10x + 25 = 0, and also determine the point of tangency.

Solution

Given the equation of the line = y = x + 5

Equation of the circle = x^2 + y^2 + 10x + 25= 0

Substitute y = x + 5 into the equation of the circle x^2 + y^2 + 10x + 25 = 0 to get the x and y intercepts of the tangent line:

x^2 + (x + 5)^2 + 10 x + 25 = 0

x^2 + x^2 + 10x + 25 + 10 x + 25 = 0

Add the like terms together to get the following expression:

2x^2 + 20x + 50 = 0

Divide the equation by 2 to simplify it:

x^2 + 10x + 25 = 0

Factorize the quadratic equation to find roots:

x^2 + 5x + 5x + 25 = 0

x (x + 5) + 5 (x + 5) = 0

(x + 5) (x + 5) = 0

x = -5

Substitute x = -5 in the equation of a line y = x + 5 to get the value of y coordinate:

y = - 5 + 5 = 0

Hence, the line y = x + 5 intersects the circle x^2 + y^2 + 10x + 25 = 0 at a single point only. We know that a line tangent to the circle intersects it at a single point only.  Therefore, we can say that the line in this example is tangent to the circle. The tangent point in a coordinate plane is (-5, 0).

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Rafia Shabbir