Let us assume that A, B and C are not collinear. Is there always a circle passing through A, B and C? Can there be more than one circle through A, B and C?
Step-by-step solution
To find: Whether a circle passes through A, B, C, and how many such circles there are
Idea: A centre O must satisfy OA = OB and OA = OC. The points with OA = OB form the perpendicular bisector of AB, and the points with OA = OC form the perpendicular bisector of AC. Because A, B, C are not collinear, these two lines meet in exactly one point, and that point is the centre.
- If a circle passes through A, B, C, let its centre be O. Then OA = OB = OC (radii).½ mark
- OA = OB, so O lies on the perpendicular bisector of AB. OA = OC, so O lies on the perpendicular bisector of AC.1 mark
- AB and AC are not along the same line (A, B, C are not collinear), so their perpendicular bisectors are not parallel. Two intersecting lines meet at exactly one point. So O can only be that point: there cannot be more than one such circle.1 mark
- Such a circle does exist: let O be the point where the two bisectors meet. Being on the first bisector, OA = OB; being on the second, OA = OC. So OA = OB = OC.1 mark
- The circle with centre O and radius OA therefore passes through A, B and C. Hence there is a unique circle through three non-collinear points.½ mark
- This circle is the circumcircle of ΔABC and O is its circumcentre. The perpendicular bisector of BC also passes through O, since OB = OC. (O lies inside an acute-angled triangle, outside an obtuse-angled triangle, and at the midpoint of the hypotenuse of a right-angled triangle.)
Answer to write in the exam
Let O be the centre of a circle through A, B, C ⇒ OA = OB = OC (radii)
OA = OB ⇒ O on the perpendicular bisector of AB; OA = OC ⇒ O on the perpendicular bisector of AC
A, B, C not collinear ⇒ the two bisectors meet in exactly one point ⇒ at most one circle
Take O = their point of intersection ⇒ OA = OB and OA = OC
∴ The circle with centre O and radius OA passes through A, B, C, and it is unique.
Common mistakes that cost marks
- Proving only that the circle exists and forgetting uniqueness (or the other way round). The question asks both.
- Forgetting why the bisectors must meet: they meet because A, B, C are not collinear; for collinear points the bisectors are parallel.
- Thinking the circumcentre is always inside the triangle. It is outside for an obtuse-angled triangle and on the hypotenuse for a right-angled one.
How this can come in the exam
The circumcentre of a triangle is the point where the
- angle bisectors of the triangle meet
- medians of the triangle meet
- perpendicular bisectors of the sides meet
- altitudes of the triangle meet
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(C) perpendicular bisectors of the sides meet
The circumcentre is equidistant from all three vertices, so it lies on the perpendicular bisector of every side.
Show that two different circles cannot meet in more than two points.
Show answer
Suppose two circles have three common points. These points cannot be collinear, because no circle passes through three collinear points (1 mark). But through three non-collinear points there is exactly one circle, so the two circles would be the same circle, which is a contradiction. Hence two different circles meet in at most two points (1 mark).Try one yourself
In ΔPQR, ∠Q = 90°. Where is the circumcentre? If PQ = 6 cm and QR = 8 cm, find the circumradius.
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For a right-angled triangle the circumcentre is the midpoint of the hypotenuse PR. PR = √(36 + 64) = 10 cm, so the circumradius is 5 cm.
More questions like this
- Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
- Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
- Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
- What is the least possible radius of a circle through two points A and B?
- 1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?