MathematicsCore18 min read

Practical Geometry of Circles

Drawing the circle a problem describes, with compasses and nothing else

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01

Every circle construction starts by locating a centre

This chapter has no new theory. It converts the circle theorems you already know into drawing instructions, and the whole of it rests on one habit: before drawing any circle, find the point that must be its centre.

The centre is always defined by a distance condition, and a distance condition is a locus. Two loci fix a point, and once you have the point, one compass setting finishes the job.

Circle wantedCentre lies onBecause
Through 3 pointsthe perpendicular bisectors of the joining chordsequidistant from two points ⟹ on their perpendicular bisector
Touching 3 sides of a trianglethe angle bisectorsequidistant from two lines ⟹ on the angle bisector
Touching a line at a given pointthe perpendicular to the line at that pointthe tangent is perpendicular to the radius
Given radius through 2 pointsarcs of that radius from each pointthe centre is r from both
02

Circumcircle and incircle

The circumcircle passes through all three vertices of a triangle. Its centre, the circumcentre, is equidistant from the three vertices, so it lies where the perpendicular bisectors of the sides meet.

The incircle touches all three sides. Its centre, the incentre, is equidistant from the three sides, so it lies where the angle bisectors meet. The radius is the perpendicular distance from that point to any side — you must drop that perpendicular to find it, not measure to a vertex.

  • Circumcentre — perpendicular bisectors of the sides. Falls outside the triangle when the triangle is obtuse, and exactly on the hypotenuse when it is right-angled.
  • Incentre — angle bisectors. Always inside the triangle, whatever its shape.
  • For a right-angled triangle the circumcircle has the hypotenuse as its diameter — which is the angle-in-a-semicircle theorem read backwards.

The Equal from 2 points option is the circumcentre construction, and Equal from 2 lines is the incentre construction. Every circle construction in this chapter is one of these two, used twice.

03

Tangent constructions

Two tangent constructions are on the syllabus, and both come straight from the tangent–radius right angle.

To draw a tangent at a point T on the circle: join OT and construct the perpendicular to OT at T. That perpendicular is the tangent.

To draw the tangents from an external point P: join OP and find its midpoint M. Draw a circle centred M with radius MP. It cuts the original circle at two points, and each of those is a point of contact. The reason is the angle in a semicircle — any point on the circle with diameter OP sees OP at 90°, which is exactly the tangent–radius right angle you need.

Worked example

Construct the tangents to a circle of radius 3 cm from a point P that is 7 cm from the centre O, and calculate their length.

  1. Draw the circle centre O, radius 3 cm, and mark P with OP = 7 cm.P is outside the circle since 7 > 3, so two tangents exist.
  2. Bisect OP to find its midpoint M, using equal arcs from O and from P.The perpendicular bisector construction gives M, and the arcs are the evidence for the mark.
  3. With centre M and radius MP = 3.5 cm, draw a circle. It cuts the first circle at A and B.Every point on this second circle sees OP at 90°, so ∠OAP = 90° and PA must be a tangent.
  4. Join PA and PB. These are the two tangents.
  5. Length: triangle OAP is right-angled at A, so PA = √(7² − 3²) = √40 = 6.32 cm.Measuring your drawing should give the same value to within a millimetre, which checks the construction.

Two tangents, each of length √40 ≈ 6.32 cm

The construction is the answer, not the picture

Marks in this chapter are for the arcs: the bisector arcs, the arcs locating the centre, the arcs of the second circle. A beautifully drawn tangent placed by eye scores nothing. Keep every arc, use a sharp pencil, and label the points you construct.

04

Checking a construction before you hand it in

Constructions are the one place in a maths paper where you can verify your own answer by measuring. Two quick checks catch nearly every error.

For a circumcircle, measure from your centre to all three vertices — the three distances must agree. For an incircle, drop a perpendicular to each side and check those three distances agree. If they do not, one of your bisectors is out and you can redraw it before losing the marks.

Before you leave this chapter

  1. Equidistant from points → perpendicular bisectors. Equidistant from lines → angle bisectors.
  2. Circumcentre = perpendicular bisectors of the sides; may fall outside the triangle.
  3. Incentre = angle bisectors; always inside. Its radius is the perpendicular distance to a side.
  4. Tangent at T: construct the perpendicular to OT at T. Tangents from P: draw the circle on OP as diameter.
  5. Leave all construction arcs visible — they are what the marks are for.
05

Circles that touch each other

Two circles that meet at exactly one point are said to touch, and the point of contact always lies on the straight line joining the two centres. That single fact answers every question in this part of the syllabus.

They can touch externally, sitting outside one another, in which case the distance between the centres is the sum of the radii. Or one can sit inside the other and touch internally, in which case the distance is the difference of the radii.

external contact: d = r₁ + r₂internal contact: d = |r₁ − r₂|intersecting at two points: |r₁ − r₂| < d < r₁ + r₂d is the distance between the two centres

Using it to construct

To draw a circle of radius 3 cm touching a given circle of radius 5 cm externally, mark a point 8 cm from the given centre and use it as the new centre. For internal contact, mark a point 2 cm away instead. The construction is a single compass arc once you have decided which of the two cases the question describes.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
How is the centre of the circle passing through three given points located?
Model answer

Join the points to form two chords and construct the perpendicular bisector of each. They meet at the circumcentre, which is equidistant from all three points.

Examiner tip. Two bisectors are enough; a third would pass through the same point and can be drawn as a check if time allows.

SQ2[2 marks]
Where is the centre of the circle that touches all three sides of a triangle, and why?
Model answer

At the incentre, where the three angle bisectors meet. A point equidistant from two lines lies on the bisector of the angle between them, so the intersection of two bisectors is equidistant from all three sides.

Examiner tip. Give the reason as well as the name. "Equidistant from two lines" is the phrase the mark scheme wants.

SQ3[2 marks]
Describe how to construct a tangent to a circle at a given point T on the circle.
Model answer

Join the centre O to T, then construct the perpendicular to OT at T. That perpendicular is the required tangent, since a tangent is always perpendicular to the radius at the point of contact.

Examiner tip. Quote the tangent–radius theorem as your justification; the construction on its own only earns half the marks.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
Construct triangle ABC with AB = 6 cm, BC = 7 cm and CA = 5 cm, then construct its circumcircle and measure its radius.
Full working
  1. Triangle constructed accurately by SSS, with arcs visibledraw the longest side first[1]
  2. Perpendicular bisector of one side constructed with equal arcs[1]
  3. Perpendicular bisector of a second side, meeting the first at the circumcentre O[1]
  4. Circle drawn through all three vertices; radius approximately 3.6 cmaccept 3.5–3.7 cm[1]

Circumradius ≈ 3.6 cm

Examiner tip. Before drawing the circle, check that your centre is the same distance from all three vertices. If two agree and one does not, redraw the bisector for the side you have not yet used.

N2[4 marks]
A point P lies 10 cm from the centre O of a circle of radius 6 cm. Construct the two tangents from P and calculate their length.
Full working
  1. Circle and point P marked with OP = 10 cmP is outside since 10 > 6[1]
  2. Midpoint M of OP found by perpendicular bisectorarcs must be visible[1]
  3. Circle centre M radius 5 cm drawn, cutting the given circle at the two points of contactthe angle in a semicircle guarantees the right angle[1]
  4. Tangent length = √(10² − 6²) = √64 = 8 cmmeasurement should agree to within 1 mm[1]

Each tangent is 8 cm long.

Examiner tip. Calculating the length as well as constructing it gives you a check on your own drawing. Here 6, 8, 10 is a Pythagorean triple, which is exactly why the examiner chose those numbers.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Triangle PQR has PQ = 8 cm, ∠PQR = 60° and QR = 6 cm.
  1. Construct the triangle.
  2. Construct its incircle.
  3. Explain why the incentre always lies inside the triangle whereas the circumcentre need not.
Mark scheme
  1. 60° angle constructed at Q with compasses, not measured with a protractorthe equilateral-triangle construction[1]
  2. PQ = 8 cm and QR = 6 cm marked off, and PR joined[1]
  3. Angle bisector of one angle constructed with arcs[1]
  4. Angle bisector of a second angle, meeting the first at the incentre I[1]
  5. Perpendicular dropped from I to a side to obtain the radius, and the circle drawn touching all three sidesmeasuring from I to a vertex is the standard error here[1]
  6. An angle bisector always lies inside the triangle, so the bisectors must meet inside; a perpendicular bisector is not confined to the interior, so in an obtuse triangle the circumcentre falls outside[1]

(c) angle bisectors are interior lines, so their intersection is interior; perpendicular bisectors are not, so the circumcentre can lie outside.

Examiner tip. The incircle radius is the perpendicular distance from the incentre to a side. Drop that perpendicular with a construction rather than estimating, or the circle will cut the sides instead of touching them.