How many circles can you draw through three distinct points A, B and C on a plane? Is there always at least one such circle? Not necessarily! What if A, B and C lie on a straight line, i.e., are collinear? Can you explain why, in this case, there is no circle through A, B and C?
Step-by-step solution
Idea: A centre must be equidistant from all three points. Points equidistant from A and B form the perpendicular bisector of AB; points equidistant from B and C form the perpendicular bisector of BC. A centre exists only if these two lines meet.
- Suppose a circle with centre O passes through A, B and C. Then OA = OB = OC (all radii).½ mark
- OA = OB, so O lies on the perpendicular bisector of AB. OB = OC, so O lies on the perpendicular bisector of BC.1 mark
- A, B, C lie on one line l. Both bisectors are perpendicular to l, so they are parallel. They are different lines, because they cross l at different points (the midpoints of AB and BC are different, since A, B, C are distinct).1 mark
- Two different parallel lines have no common point, so no point O can exist. Hence no circle passes through three collinear points.½ mark
Answer to write in the exam
Suppose a circle with centre O passes through A, B, C, which lie on line l.
OA = OB ⇒ O lies on the perpendicular bisector of AB
OB = OC ⇒ O lies on the perpendicular bisector of BC
Both bisectors ⟂ l ⇒ they are parallel, and distinct (different midpoints)
∴ No such O exists; no circle passes through three collinear points.
Common mistakes that cost marks
- Saying “a circle is curved, a line is straight” without a reason. The proof must show that no centre can exist.
- Forgetting to say the two bisectors are different lines. Parallel lines that are the same line would meet everywhere.
- Thinking a very large circle can pass through three collinear points. However large, a circle meets a straight line in at most two points.
How this can come in the exam
How many circles pass through three points that lie on one straight line?
- 0
- 1
- 2
- Infinitely many
Show answer
(A) 0
The perpendicular bisectors of the two segments are parallel, so there is no point equidistant from all three points: no centre, no circle.
Try one yourself
P(1, 2), Q(3, 2) and R(7, 2) are three points. Can a circle pass through all three? Explain using perpendicular bisectors.
Show answer
No. The three points lie on the line y = 2. The perpendicular bisector of PQ is the line x = 2 and that of QR is x = 5. These are parallel and never meet, so no point is equidistant from P, Q and R.
More questions like this
- 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?
- 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?