Circle-line intersection

If we want to find the points of intersection between a circle and a straight line, there are three possibilities:

• The line does not intersect the circle (0 points of intersection)
• The line is tangent to the circle (1 point of intersection)
• The line intersects the circle in two places (2 points of intersection)

See number of intersections between graphs

Finding the points of intersection between a circle and a line

If we want to find the points of intersection, we can use good old simultaneous equations! We won’t be able to use elimination, but we can use substitution.

Here are the steps:

1. Write down the equations of the circle and the line.
2. Rearrange the equation of the line to make y the subject (if it isn’t already).
3. Substitute the expression for y from the line equation into the circle equation.
4. Rearrange the resulting equation into the form ax^2 + bx + c = 0.
5. Solve the quadratic equation using the quadratic formula, factorising, or completing the square.
6. Substitute the x-values found back into the equation of the line to find the corresponding y-values.

That’s sounds like a lot of steps, but let’s see it in practice with an example.

find the points of intersection between the circle x^2 + y^2 = 25 and the line y = 2x + 1.

• Write down the equations:
  • Circle: x^2 + y^2 = 25
  • Line: y = 2x + 1
• Substitute the expression for y from the line equation into the circle equation:
  • x^2 + (2x + 1)^2 = 25
• Expand the equation:
  • x^2 + (4x^2 + 4x + 1) = 25
  • x^2 + 4x^2 + 4x + 1 = 25
  • 5x^2 + 4x + 1 = 25
• Rearrange into the form ax^2 + bx + c = 0:
  • 5x^2 + 4x + 1 - 25 = 0
  • 5x^2 + 4x - 24 = 0
• Solve the quadratic equation using the quadratic formula:
  • x = \frac{-4 \pm \sqrt{4^2 - 4 \times 5 \times -24}}{2 \times 5}
  • = \frac{-4 \pm \sqrt{16 + 480}}{10}
  • = \frac{-4 \pm \sqrt{496}}{10}
  • = \frac{-4 \pm 4\sqrt{31}}{10}
  • = \frac{-2 \pm 2\sqrt{31}}{5}
• Find the corresponding y-values by substituting the x-values back into the equation of the line:
  • For x = \frac{-2 + 2\sqrt{31}}{5}:
    • y = 2(\frac{-2 + 2\sqrt{31}}{5}) + 1
    • = \frac{-4 + 4\sqrt{31}}{5} + 1
    • = \frac{-4 + 4\sqrt{31}}{5} + \frac{5}{5}
    • = \frac{1 + 4\sqrt{31}}{5}
  • For x = \frac{-2 - 2\sqrt{31}}{5}:
    • y = 2(\frac{-2 - 2\sqrt{31}}{5}) + 1
    • = \frac{-4 - 4\sqrt{31}}{5} + 1
    • = \frac{-4 - 4\sqrt{31}}{5} + \frac{5}{5}
    • = \frac{1 - 4\sqrt{31}}{5}
• Answer: The points of intersection are:
  • \frac{-2 + 2\sqrt{31}}{5}, \frac{1 + 4\sqrt{31}}{5})
  • \frac{-2 - 2\sqrt{31}}{5}, \frac{1 - 4\sqrt{31}}{5})