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

Intersection of a Line and a Circle

When we work with coordinate geometry, we often need to find where a straight line meets a circle. This intersection can happen in different ways, and understanding these patterns is crucial for solving problems effectively.

Understanding the types of intersection

A line can interact with a circle in three possible ways:

No intersection - the line misses the circle completely
Two intersection points - the line crosses through the circle
One intersection point - the line just touches the circle (this is called a tangent)

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Key definition: When a line intersects a circle at exactly one point, that line is called a tangent to the circle. The single point where they meet is called the point of contact.

General method for finding intersection points

To find where a line and circle intersect, we use simultaneous equations. Here's the step-by-step approach:

Express the line equation in the form $y = ...$ or $x = ...$ (choose whichever avoids fractions if possible)
Substitute this expression into the circle equation
Solve the resulting equation to find the coordinate values
Calculate the intersection points by substituting back

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The number of solutions tells us about the relationship:

Two different solutions = line intersects circle at two points
One repeated solution = line is tangent to circle
No real solutions = line doesn't intersect circle

Worked example 1: Two intersection points

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Worked Example: Finding Two Intersection Points

Find the intersection points of the line $x + 3y - 5 = 0$ and the circle $x^2 + y^2 = 5$ .

Step 1: Express $x$ in terms of $y$ from the line equation $x + 3y - 5 = 0$ Therefore: $x = -3y + 5$

Step 2: Substitute this into the circle equation $x^2 + y^2 = 5$ $(-3y + 5)^2 + y^2 = 5$

Step 3: Expand and simplify $9y^2 - 30y + 25 + y^2 = 5$ $10y^2 - 30y + 25 = 5$ $10y^2 - 30y + 20 = 0$

Step 4: Divide by 10 to simplify $y^2 - 3y + 2 = 0$

Step 5: Factorise $(y - 2)(y - 1) = 0$ So $y = 2$ or $y = 1$

Step 6: Find the $x$ -coordinates by substituting back When $y = 2$ : $x = -3(2) + 5 = -1$ , giving point $(-1, 2)$ When $y = 1$ : $x = -3(1) + 5 = 2$ , giving point $(2, 1)$

Answer: The intersection points are $(-1, 2)$ and $(2, 1)$ .

Worked example 2: Tangent line

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Worked Example: Proving a Line is Tangent

Show that the line $3x + y + 10 = 0$ is tangent to the circle $x^2 + y^2 = 10$ , and find the point of contact.

Step 1: Express $y$ in terms of $x$ (to avoid fractions) $3x + y + 10 = 0$ Therefore: $y = -3x - 10$

Step 2: Substitute into the circle equation $x^2 + (-3x - 10)^2 = 10$

Step 3: Expand and simplify $x^2 + 9x^2 + 60x + 100 = 10$ $10x^2 + 60x + 100 = 10$ $10x^2 + 60x + 90 = 0$

Step 4: Divide by 10 $x^2 + 6x + 9 = 0$

Step 5: Factorise $(x + 3)(x + 3) = 0$ This gives $x = -3$ (repeated solution)

Step 6: Find $y$ -coordinate When $x = -3$ : $y = -3(-3) - 10 = 9 - 10 = -1$

Answer: The point of contact is $(-3, -1)$ . Since there's only one solution, the line is indeed tangent to the circle.

Exam tips and common mistakes

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Avoid These Common Mistakes:

Choose your substitution wisely - express the line equation to avoid fractions where possible
Check your working - if you get one repeated solution, the line is tangent
Be careful with signs - negative coordinates are common in these problems
Verify your answers - substitute your points back into both original equations to check

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Key Points to Remember:

A tangent line intersects a circle at exactly one point
Use simultaneous equations by substituting the line equation into the circle equation
Two solutions = two intersection points, one repeated solution = tangent line
Always express the line equation in the simplest form before substituting
Check your final answers by substituting back into the original equations

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