Find the equation of a tangent at a circle point
Use the radius as a normal to the tangent and write the line through the contact point.
Section 8 · Coordinate geometry of the circle
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the angle between a radius and a tangent at their contact point?
What you will learn
- Recognise that a tangent is perpendicular to the radius at contact
- Find a tangent gradient from the radius gradient
- Write and check the tangent equation at a stated point
The radius meets its tangent at 90°
At a point T on a circle, the radius from centre C to T is perpendicular to the tangent. The radius gradient is the normal direction to the tangent.
For finite non-zero radius gradient mᵣ, the tangent gradient is −1/mᵣ. Then use point-gradient form through T.
Check that T satisfies the circle equation and the tangent equation. A horizontal or vertical radius has a vertical or horizontal tangent respectively.
The centre and contact point determine the tangent
First verify the point lies on the circle. Find the radius vector by subtracting centre coordinates from contact-point coordinates.
If the radius gradient is undefined, the radius is vertical and the tangent is horizontal. If the radius is horizontal, its tangent is vertical; avoid dividing by zero.
Pause and explain
What is the tangent gradient when the radius gradient is −2?
Worked example
Find the tangent to (x − 2)² + (y + 1)² = 25 at (5, 3).
Show the worked solution
- The centre is (2, −1); the radius vector to (5, 3) is (3, 4), so its gradient is 4/3.
- The tangent gradient is −3/4.
- Through (5, 3), the tangent is y − 3 = −(3/4)(x − 5), or 3x + 4y = 27.
Answer y − 3 = −(3/4)(x − 5), equivalently 3x + 4y = 27.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the tangent to x² + y² = 25 at (3, 4).
Give me a hint
The centre is the origin; find the radius gradient and then its perpendicular.
Compare my reasoning
- The radius gradient is 4/3.
- The tangent gradient is −3/4.
- Through (3, 4), y − 4 = −(3/4)(x − 3), so 3x + 4y = 25.
3x + 4y = 25.
Look for these in your work
- Point lies on radius 5
- Perpendicular gradient used
- Tangent passes through contact point
Find the tangent to the circle centred at (−1, 2) through its point (3, 5).
Give me a hint
Use the centre-to-point vector (4, 3).
Compare my reasoning
- The radius gradient is 3/4.
- The tangent gradient is −4/3.
- Through (3, 5), y − 5 = −(4/3)(x − 3), so 4x + 3y = 27.
4x + 3y = 27.
Look for these in your work
- Radius direction calculated
- Negative reciprocal used
- Equation contains the contact point
Find the tangent at (0, 3) to x² + y² = 9.
Give me a hint
The radius from (0, 0) to (0, 3) is vertical, so the tangent is horizontal.
Compare my reasoning
- The centre and contact point share x = 0, so the radius is vertical.
- A horizontal line through (0, 3) has equation y = 3.
- The point lies on the circle because 0² + 3² = 9.
y = 3.
Look for these in your work
- Vertical radius recognised
- Horizontal tangent stated
- Circle point checked
Common mistakes
- Using the radius gradient as the tangent gradient
- Taking a reciprocal but keeping the same sign
- Finding a line that misses the stated contact point
- Trying to use a finite gradient for a vertical line
What can you explain now?
A radius has gradient 4/3. What is the tangent gradient at its endpoint?
Compare with the explanation
−3/4
The tangent is perpendicular, so the gradients multiply to −1.
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Make it stick
How does the centre-to-contact-point line determine the tangent equation?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.