Construct both tangents from an external point
Construct the auxiliary circle with a diameter OP · Locate tangent points using a semicircle right angle · Distinguish external, on-circle and internal points
Section Paper 4 · Historical Extended · calculator
Exam scope and assessment notes
Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Perpendicular bisectors and circle definitions
- Right-triangle relation a²+b²=c²
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Construct the auxiliary circle with a diameter OP
- Locate tangent points using a semicircle right angle
- Distinguish external, on-circle and internal points
Prepare your paper
When a practice task gives numerical input geometry rather than a printed figure, first prepare that source segment, angle, triangle or circle with measuring tools as needed. Then put the measuring scale or protractor aside for the requested formal copying or bisector steps: use a compass and unmarked straightedge to locate the new points. The preparation supplies the givens; it is not a replacement for the formal construction.
The website diagrams explain relationships; their screen size is not a centimetre calibration or a ready-to-measure template. Keep the input figure and useful construction arcs on paper. Your drawing is self-checked and needs human review for reliable marking.
A circle that locates a right angle
Given circle center O and external point P, join OP and construct its midpoint M. Draw an auxiliary circle centered at M with radius MO, so OP is its diameter. Its two intersections T₁/T₂ with the given circle locate the tangent points. Draw PT₁ and PT₂.
Each angle OT₁P or OT₂P is 90° because its vertex is on the auxiliary circle with diameter OP. Thus each line PT is perpendicular to the given circle's radius OT at contact and is a tangent. Keep the auxiliary circle or its relevant arcs visible.
Check existence and length
This two-tangent method requires OP greater than the given radius r. For an on-circle point there is one tangent, constructed perpendicular to its radius at that point; for an internal point there is no real tangent line from that point to the circle.
A right triangle OTP gives PT² = OP²−r². In the illustrated r = 3, OP = 5 case, each tangent has length √(25−9) = 4. The equal lengths are a check, not a substitute for locating the tangent points formally.
Sources for this lesson
Pause and explain
Why does the diameter-OP circle locate tangent points?
Worked example
Construct tangents from P when OP = 5 and the circle radius is 3.
Show the worked solution
- Construct midpoint M and the circle with diameter OP.
- Join P to each of its two intersections with the original circle.
- Each contact radius is perpendicular to its tangent; the length check gives 4.
Answer Two tangents, each length 4, located using the diameter-OP circle.
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.
For radius 3 and OP = 5, describe the tangent construction and check each tangent length.
Give me a hint
Locate T before using the right-triangle calculation.
Compare my reasoning
- Construct the OP midpoint and auxiliary diameter circle, then join P to its two circle intersections.
- PT² = 5²−3² = 16, so PT = 4.
Two formally located tangents; each has length 4.
Look for these in your work
- Retain the auxiliary-circle evidence
- Use the perpendicular radius condition
A circle has radius 5 and an external point has OP = 13. Find the tangent length and state why two contacts exist.
Give me a hint
Use OP > r and the right triangle.
Compare my reasoning
- 13 > 5 gives two real tangent contacts.
- PT = √(169−25) = 12.
Two tangents, each length 12.
Look for these in your work
- Check external location
- Select the positive length
For a radius-3 circle, compare points at center distances 2, 3 and 5. How many tangent lines from each?
Give me a hint
Separate inside, on and outside cases.
Compare my reasoning
- Distance 2 is inside: none. Distance 3 is on-circle: one perpendicular to the radius.
- Distance 5 is outside: two, using the auxiliary circle.
0, 1 and 2 tangents, respectively.
Look for these in your work
- Check the geometric domain
- Do not apply the external construction unchanged at contact
Common mistakes
- The line OP itself is not a tangent.
- The external-point assumption cannot be dropped.
What can you explain now?
Why does the diameter-OP circle locate tangent points?
Compare with the explanation
It makes each contact angle OTP a right angle.
A tangent is perpendicular to the contact radius.
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Make it stick
Why does the diameter-OP circle locate tangent points?
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.