Find a circle tangent from its perpendicular radius
Check the point lies on the circle, then use the radius direction to construct a tangent without calculus.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use algebra and elementary geometry; implicit differentiation is outside this section. Written practice is self-checked.
Before you begin
- Find a circle's centre and check a point in its equation.
- Build perpendicular lines from points and gradients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which radius gradient connects C(2, −1) to P(5, 3)?
What you will learn
- Use a tangent perpendicular to the radius at the point of contact.
- Handle horizontal and vertical tangents and intersections of two tangents.
Begin with the centre and contact point
A circle's tangent at P is perpendicular to the radius CP and passes through P. First verify that P lies on the circle; a line through an arbitrary point with a perpendicular direction is not automatically a tangent there. For centre C(2, −1) and P(5, 3), the squared radius distance is 3² + 4² = 25, so P belongs to the circle (x − 2)² + (y + 1)² = 25.
The radius gradient is (3 + 1)/(5 − 2) = 4/3, hence the tangent gradient is −3/4. Through P, its equation is y − 3 = −3(x − 5)/4, or 3x + 4y = 27. This uses the radius-tangent theorem and line equations; implicit differentiation is not part of this section's method.
Use geometry for exceptional directions
If a radius is vertical, the tangent is horizontal. If a radius is horizontal, the tangent is vertical. For a circle with centre (−1, 2) and radius 5, its top point (−1, 7) has tangent y = 7. A reciprocal-gradient calculation is unnecessary and undefined for a horizontal radius, so handle the geometry directly.
Two tangents from the same exterior point to a circle have equal lengths. One way to check is to use the two right triangles formed by the common centre and the contact points: they have the same centre-to-exterior distance and equal radii. If the contact points are supplied, construct both tangent equations, solve them simultaneously and then calculate the distances to verify the result.
Pause and explain
Which line is the tangent to (x − 2)² + (y + 1)² = 25 at (5, 3)?
Worked example
Find the tangent to (x − 2)² + (y + 1)² = 25 at P(5, 3).
Show the worked solution
- The centre is C(2, −1), and CP² = 3² + 4² = 25, confirming P lies on the circle.
- The radius gradient is 4/3; its perpendicular tangent gradient is −3/4.
- Use y − 3 = −3(x − 5)/4 and rearrange to 3x + 4y = 27.
Answer 3x + 4y = 27.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 its negative reciprocal.
Compare my reasoning
- The point satisfies 3² + 4² = 25, and the radius gradient is 4/3.
- The tangent has gradient −3/4 through (3, 4).
- Write y − 4 = −3(x − 3)/4 and simplify to 3x + 4y = 25.
3x + 4y = 25.
Look for these in your work
- Verify the contact point lies on the circle.
- Use a perpendicular gradient and the correct point.
A circle has centre (−1, 2) and radius 5. Find its tangent at (−1, 7).
Give me a hint
The contact point and centre have the same x-coordinate.
Compare my reasoning
- The point is 5 units above the centre, so it lies on the circle.
- The radius is vertical, making the tangent horizontal.
- The horizontal line through the contact point is y = 7.
y = 7.
Look for these in your work
- Recognise the exceptional vertical-radius case.
- Position the horizontal tangent through the contact point.
Tangents to x² + y² = 25 at (3, 4) and (−3, 4) meet at T. Find T and both tangent lengths.
Give me a hint
Build both tangent equations and solve them together.
Compare my reasoning
- The equations are 3x + 4y = 25 and −3x + 4y = 25.
- Subtract to get x = 0, then y = 25/4, giving T = (0, 25/4).
- Each tangent length is √(3² + (25/4 − 4)²) = √(9 + 81/16) = 15/4; the equal lengths also follow from the tangent theorem.
T = (0, 25/4); both tangent lengths are 15/4.
Look for these in your work
- Construct a valid equation for each tangent.
- Find the intersection and verify equal lengths using distance.
Common mistakes
- The perpendicular radius is drawn to the contact point, not to an arbitrary point on the line.
- Check that the given contact point is on the circle.
- Use horizontal or vertical line equations in the exceptional gradient cases.
What can you explain now?
For the circle x² + y² = 25, find the tangent at its rightmost point (5, 0).
Compare with the explanation
x = 5.
The radius from (0, 0) to (5, 0) is horizontal, so its perpendicular tangent is the vertical line through (5, 0).
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain why a tangent must pass through the contact point and be perpendicular to its radius. Construct examples with both finite and undefined radius gradients.
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.