Intersection of a Line and a Circle Revision Notes for Leaving Cert Mathematics

Intersection of a Line and a Circle

What is the Intersection of a Line and a Circle?

The intersection of a line and a circle can result in:

No Intersection: The line does not touch the circle.
One Point of Intersection: The line is tangent to the circle.
Two Points of Intersection: The line crosses through the circle.

Determining the Intersection Points

To find the points of intersection between a line and a circle:

Substitute the Equation of the Line into the Circle:

Equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$
Equation of a line: $y = mx + c$

Solve the Resulting Quadratic Equation:

Substitute $y = mx + c$ into $(x - h)^2 + (y - k)^2 = r^2$
This gives a quadratic equation in $x$

Analyse the Discriminant ( $\Delta$ ):

$\Delta = b^2 - 4ac$ , where $a, b, c$ are coefficients from the quadratic equation.
$\Delta > 0$ : Two points of intersection.
$\Delta = 0$ : One point of intersection (tangent).
$\Delta < 0$ : No intersection.

Worked Examples

infoNote

Example 1: Find Points of Intersection

Problem: Find the points of intersection between the circle $x^2 + y^2 = 25$ and the line $y = 2x - 1$ .

Solution:

Step 1: Substitute $y = 2x - 1$ into $x^2 + y^2 = 25$ :

x^2 + (2x - 1)^2 = 25

Step 2: Expand and simplify:

x^2 + 4x^2 - 4x + 1 = 25

5x^2 - 4x - 24 = 0

Step 3: Solve the quadratic equation using the quadratic formula:

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(5)(-24)}}{2(5)} = \frac{4 \pm \sqrt{16 + 480}}{10}

x = \frac{4 \pm \sqrt{496}}{10} = \frac{4 \pm 4\sqrt{31}}{10} = \frac{2 \pm 2\sqrt{31}}{5}

Step 4: Substitute $x$ values back into $y = 2x - 1$ to find $y$ coordinates:

y_1 = 2\left(\frac{2 + 2\sqrt{31}}{5}\right) - 1

y_2 = 2\left(\frac{2 - 2\sqrt{31}}{5}\right) - 1

Answer: Points of intersection are:

\left(\frac{2 + 2\sqrt{31}}{5}, \frac{4 + 4\sqrt{31}}{5} - 1\right), \quad \left(\frac{2 - 2\sqrt{31}}{5}, \frac{4 - 4\sqrt{31}}{5} - 1\right)

infoNote

Example 2: Determine Number of Intersections

Problem: Show that the line $y = -\frac{1}{2}x + 3$ intersects the circle $x^2 + y^2 = 10$ at two points.

Solution:

Step 1: Substitute $y = -\frac{1}{2}x + 3$ into $x^2 + y^2 = 10$ :

x^2 + \left(-\frac{1}{2}x + 3\right)^2 = 10

Step 2: Expand and simplify:

x^2 + \frac{1}{4}x^2 - 3x + 9 = 10

\frac{5}{4}x^2 - 3x - 1 = 0

Step 3: Calculate the discriminant ( $\Delta$ ):

\Delta = (-3)^2 - 4\left(\frac{5}{4}\right)(-1) = 9 + 5 = 14

Answer: The discriminant is $14$ , so there are two points of intersection.

Summary

Steps for Finding Intersection:
1. Substitute the line's equation into the circle's equation.
2. Solve the resulting quadratic equation.
3. Use the discriminant ( $\Delta$ ) to classify the intersection.
Key Cases:
- $\Delta > 0$ : Two points of intersection.
- $\Delta = 0$ : Tangency (one point).
- $\Delta < 0$ : No intersection.
Practice solving such problems to understand tangency and intersections with circles.

Intersection of a Line and a Circle (Leaving Cert Mathematics): Revision Notes