Classify line-circle intersections with substitution
Turn common points into an equation and use its real solutions to distinguish chords, tangency and no intersection.
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
- Use circle equations and solve quadratic equations.
- Interpret a discriminant and solve inequalities involving a square.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Substitute y = x + 2 into x² + y² = 10. Which equation follows?
What you will learn
- Find exact intersections of a line and a circle.
- Find parameter conditions for two intersections, tangency or no real intersections.
Choose the substitution to suit the line
A common point satisfies both the circle and line equations. For a non-vertical line y = mx + c, substitute this expression for every y in the circle, keeping brackets when squaring. Solve the resulting quadratic in x, then recover the y-coordinate for each root. A vertical line x = t instead supplies x directly and leaves an equation in y.
For (x − 2)² + (y + 1)² = 25 and x = t, the intersection equation is (y + 1)² = 25 − (t − 2)². Two points occur when the right side is positive; zero gives one point; a negative value gives none. Thus two intersections require −3 < t < 7, tangency occurs at t = −3 or 7, and otherwise the line misses the circle.
The discriminant gives the same geometric classification
For a genuine intersection quadratic, D > 0 gives two distinct points, D = 0 gives a tangent and D < 0 gives no real common points. The circle's centre-to-line distance gives an equivalent interpretation: less than the radius means a chord, equal means a tangent, and greater means no intersection. You can use elementary geometry or substitution rather than assuming a point count from a rough sketch.
For x² + y² = 10 and y = x + k, substitution gives 2x² + 2kx + k² − 10 = 0. Its discriminant is 80 − 4k². Therefore the line meets twice when |k| < 2√5, touches when |k| = 2√5 and misses when |k| > 2√5. At tangency, x = −k/2 and y = k/2. Include both signs when solving a condition on k².
Pause and explain
What is the line-circle geometry when the genuine intersection quadratic has D < 0?
Worked example
Classify the intersections of x = t with (x − 2)² + (y + 1)² = 25 for real t.
Show the worked solution
- Substitute x = t to obtain (y + 1)² = 25 − (t − 2)².
- Two real y-values require (t − 2)² < 25, or −3 < t < 7.
- Equality gives t = −3 or 7 with y = −1; outside those limits the squared quantity would need to be negative.
Answer Two intersections for −3 < t < 7; tangent at (−3, −1) or (7, −1) for t = −3 or 7; none for t < −3 or t > 7.
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 both intersections of y = 3 with x² + y² = 25.
Give me a hint
Replace y by 3, then include both signs when solving for x.
Compare my reasoning
- Substitute to get x² + 9 = 25.
- Thus x² = 16, giving x = −4 or 4.
- Pair each root with the common y-coordinate 3 and check the circle equation.
(−4, 3) and (4, 3).
Look for these in your work
- Retain both square-root branches.
- State both complete intersection coordinates.
Find all intersections of y = x + 2 with x² + y² = 10.
Give me a hint
Expand the substituted square before factorising.
Compare my reasoning
- Write x² + (x + 2)² = 10, giving 2x² + 4x − 6 = 0.
- Divide by 2 and factor (x + 3)(x − 1) = 0, so x = −3 or 1.
- The corresponding line outputs are −1 and 3; both pairs satisfy the circle.
(−3, −1) and (1, 3).
Look for these in your work
- Expand the cross term correctly.
- Recover and verify both y-coordinates.
For x² + y² = 10 and y = x + k, find all k giving two intersections, tangency or none. State the tangent point in terms of k.
Give me a hint
Apply the discriminant to 2x² + 2kx + k² − 10 = 0.
Compare my reasoning
- The discriminant is (2k)² − 8(k² − 10) = 80 − 4k².
- It is positive for k² < 20, zero for k² = 20 and negative for k² > 20.
- At either tangent value k = ±2√5, the repeated root is x = −k/2 and the line gives y = k/2.
Two for |k| < 2√5; tangent for k = ±2√5 at (−k/2, k/2); none for |k| > 2√5.
Look for these in your work
- Classify strict and equality cases from the correct discriminant.
- Retain both tangent parameter values and recover their coordinates.
Common mistakes
- A vertical line is substituted as x = constant, rather than forced into y = mx + c.
- Recover the second coordinate for every root.
- Include both signs in a squared parameter condition and keep tangent equality separate.
What can you explain now?
Find the common point of y = 5 and x² + y² = 25, and classify the line.
Compare with the explanation
(0, 5); y = 5 is a tangent.
Substitution gives x² = 0, one repeated real root, so there is one distinct common point.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why a line-circle substitution can give two roots, a repeated root or no real roots. Derive one parameter condition and connect it to a geometric sketch.
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.