Find tangents from an external point
Represent a line through an external point with an unknown gradient, then require a repeated circle intersection.
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
How many tangents can usually be drawn from a point outside a circle?
What you will learn
- Form a line through a stated external point
- Use a zero discriminant to impose tangency
- Find and interpret both tangent equations when two exist
Tangency means exactly one repeated intersection
A line through an external point generally meets a circle at zero, one or two points. A tangent has one repeated intersection, so substituting it into the circle gives a quadratic with discriminant zero.
For a circle x² + y² = r² and an external point (p, 0), write a non-vertical candidate line as y = m(x − p). Substitute and set the discriminant to zero to find tangent gradients.
A vertical line through (p, 0) is x = p. Check it separately if appropriate; for p > r it misses the circle and is not a tangent.
Check each line against the point and circle
An external point outside a circle normally has two tangents. The two gradients have opposite signs when the point lies on the positive x-axis.
Verify that each equation passes through the external point. Its circle intersection must be a repeated root, not two distinct roots.
Pause and explain
A substituted line gives a quadratic with two distinct real roots. Is the line tangent?
Worked example
Find the tangents from (13, 0) to x² + y² = 25.
Show the worked solution
- Let y = m(x − 13). Substitution gives (1 + m²)x² − 26m²x + 169m² − 25 = 0.
- Set its discriminant to zero: 100 − 576m² = 0, so m = ±5/12.
- The two tangents are y = (5/12)(x − 13) and y = −(5/12)(x − 13).
Answer y = ±(5/12)(x − 13).
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 tangents from (5, 0) to x² + y² = 9.
Give me a hint
Set y = m(x − 5), substitute and use the zero-discriminant condition.
Compare my reasoning
- The substituted quadratic is (1 + m²)x² − 10m²x + 25m² − 9 = 0.
- Its discriminant condition simplifies to 9 − 16m² = 0, so m = ±3/4.
- The tangent lines are y = ±(3/4)(x − 5).
y = (3/4)(x − 5) and y = −(3/4)(x − 5).
Look for these in your work
- Line form passes through point
- Repeated-root condition applied
- Both tangents stated
Find the tangent gradients and equations from (5, 0) to x² + y² = 16.
Give me a hint
Use the same substitution and discriminant-zero method.
Compare my reasoning
- For radius 4 and p = 5, m² = r²/(p² − r²) = 16/9.
- Thus m = ±4/3.
- The two lines are y = ±(4/3)(x − 5).
y = (4/3)(x − 5) and y = −(4/3)(x − 5).
Look for these in your work
- Gradient magnitude found
- Both signs retained
- Lines pass through (5, 0)
Find the tangent gradients from (13, 0) to a circle of radius 5 centred at the origin, and explain why there are two tangents.
Give me a hint
Tangency gives m² = r²/(p² − r²).
Compare my reasoning
- m² = 25/(169 − 25) = 25/144.
- So m = 5/12 or m = −5/12.
- Both signs give one tangent on each side of the centre line.
Gradients ±5/12; the two tangent lines are y = ±(5/12)(x − 13).
Look for these in your work
- Exact gradients calculated
- Two signs interpreted
- Both tangent equations recovered
Common mistakes
- Using a line that does not pass through the external point
- Setting the quadratic coefficient to zero instead of the discriminant
- Keeping only one of the two tangent gradients
- Accepting a secant with a positive discriminant
What can you explain now?
What discriminant condition identifies a tangent after substitution?
Compare with the explanation
The discriminant is zero
A tangent has one repeated real intersection with the circle.
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Make it stick
Why does setting a substituted quadratic's discriminant to zero find a tangent?
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.