Use circle symmetry, chords and diameters in coordinate problems
Combine a perpendicular chord bisector, Pythagoras and the angle in a semicircle with algebraic checks.
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 midpoints, exact distances and a circle's centre.
- Use Pythagoras and identify perpendicular lines.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the perpendicular distance from (2, −1) to the horizontal line y = 2?
What you will learn
- Use circle symmetry to locate a chord midpoint and calculate its length.
- Use diameter endpoints and the right angle in a semicircle to solve coordinate problems.
A centre-to-chord perpendicular halves the chord
The perpendicular from a circle's centre to a chord passes through the chord midpoint. With radius r and perpendicular centre-to-chord distance d, a right triangle has half-chord length √(r² − d²), so the full chord length is 2√(r² − d²). This requires 0 ≤ d < r for a chord with two distinct endpoints. The equality d = r gives a tangent point rather than a positive-length chord.
For centre (2, −1), radius 5 and horizontal chord y = 2, the perpendicular meets the chord at (2, 2), so d = 3. The half-chord is √(25 − 9) = 4. The endpoints are (−2, 2) and (6, 2), and the full length is 8. Direct substitution into (x − 2)² + (y + 1)² = 25 verifies both endpoints.
A diameter determines a right angle at the circumference
If A and B are diameter endpoints and P is another point on their circle, angle APB is 90°. It is the angle at P, not the angle at the centre, and P must be distinct from A and B. Use the diameter midpoint as centre and half its length as radius to verify whether a proposed P is on that circle.
For A(−2, 1) and B(4, 5), the circle is (x − 1)² + (y − 3)² = 13. Point P(4, 1) lies on it because 3² + (−2)² = 13. PA is horizontal and PB vertical, confirming the semicircle right angle without dividing by an undefined gradient. Their lengths 6 and 4 also give PA² + PB² = 52 = AB².
Pause and explain
A radius is 5 and the centre-to-chord perpendicular distance is 3. What is the full chord length?
Worked example
For (x − 2)² + (y + 1)² = 25, find the endpoints, midpoint and length of the chord on y = 2.
Show the worked solution
- The centre is (2, −1), so the perpendicular reaches y = 2 at midpoint (2, 2), a distance 3 from the centre.
- The half-chord length is √(5² − 3²) = 4, giving x = 2 − 4 or 2 + 4.
- The endpoints are (−2, 2) and (6, 2); their midpoint is (2, 2) and their separation is 8.
Answer Endpoints (−2, 2), (6, 2); midpoint (2, 2); chord length 8.
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.
A circle has radius 5. A chord is a perpendicular distance 3 from its centre. Find its length.
Give me a hint
Use the right triangle formed by the radius and half the chord.
Compare my reasoning
- The radius is the hypotenuse and the perpendicular distance is one shorter side.
- Pythagoras gives half-length √(5² − 3²) = 4.
- Symmetry gives two equal halves, so the chord length is 8.
8.
Look for these in your work
- Use the radius as the hypotenuse.
- Double the computed half-length.
Find the endpoints and length of the chord y = 6 on x² + y² = 100.
Give me a hint
Substitute the chord height into the circle equation.
Compare my reasoning
- The equation becomes x² + 36 = 100, so x² = 64.
- The endpoints are (−8, 6) and (8, 6).
- Their horizontal separation is 16; the midpoint is (0, 6), consistent with the circle's symmetry.
Endpoints (−8, 6), (8, 6); length 16.
Look for these in your work
- Include both signs for the endpoint inputs.
- Find full separation rather than the distance from midpoint to one end.
A circle has diameter endpoints A(−2, 1) and B(4, 5). Verify that P(4, 1) lies on it and explain angle APB.
Give me a hint
Find the diameter midpoint and squared radius, then use the semicircle theorem or perpendicular directions.
Compare my reasoning
- The centre is (1, 3) and r² = (6² + 4²)/4 = 13.
- For P, the squared centre distance is (4 − 1)² + (1 − 3)² = 13, so P lies on the circle and differs from both endpoints.
- Because AB is a diameter, angle APB is 90°; PA horizontal and PB vertical independently confirm it.
P lies on (x − 1)² + (y − 3)² = 13; angle APB = 90°.
Look for these in your work
- Verify the point in the actual diameter circle.
- Identify the right angle at the circumference point P.
Common mistakes
- The radius is the hypotenuse of the half-chord triangle.
- Double the half-chord length to get the full chord.
- An angle in a semicircle is at a circumference point distinct from the diameter endpoints.
What can you explain now?
A circle has radius 13 and a chord whose perpendicular distance from the centre is 5. Find the chord length.
Compare with the explanation
24.
Its half-length is √(169 − 25) = 12, so the full chord is 24.
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Make it stick
Explain the two different midpoint ideas in a diameter and a chord. Sketch the half-chord right triangle and locate the angle in a semicircle.
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.