What is the least possible radius of a circle through two points A and B?
Step-by-step solution
Idea: The centre O of any circle through A and B lies on the perpendicular bisector of AB. If M is the midpoint of AB, then OA is the hypotenuse of right triangle OMA, so OA ≥ AM, with equality only when O = M.
- Let a circle with centre O pass through A and B, and let M be the midpoint of AB. Since OA = OB, O lies on the perpendicular bisector of AB, so OM ⟂ AB (or O = M).½ mark
- In right triangle OMA, OA2 = AM2 + OM2 ≥ AM2. So the radius OA ≥ AM = ½AB.1 mark
- Equality holds when OM = 0, that is, O = M. Then AB is a diameter. So the least possible radius is ½AB.½ mark
Check: A and B 8 cm apart: centres 0, 3 and 6 cm from M give radii 4, 5 and √52 ≈ 7.2 cm. None is smaller than 4 cm = ½AB.
Answer to write in the exam
Let O be the centre of a circle through A and B; M = midpoint of AB.
OA = OB ⇒ O lies on the perpendicular bisector of AB ⇒ OM ⟂ AB
OA2 = AM2 + OM2 ≥ AM2 (Baudhāyana–Pythagoras theorem)
⇒ OA ≥ AM = ½AB, with equality when O = M
∴ Least possible radius = ½AB (circle with AB as diameter)
Common mistakes that cost marks
- Answering AB. A circle of radius AB centred at A passes through B but not through A itself; the smallest circle through both has AB as its diameter.
- Saying “any radius is possible”. A circle through A and B must have a chord of length AB, and no chord is longer than the diameter, so 2r ≥ AB.
- Forgetting the reason: OA is the hypotenuse of a right triangle with leg AM, so OA ≥ AM.
How this can come in the exam
P and Q are 9 cm apart. A circle of radius r passes through P and Q. Which must be true?
- r = 4 cm
- r ≥ 4.5 cm
- r ≤ 4.5 cm
- r = 9 cm
Show answer
(B) r ≥ 4.5 cm
PQ is a chord, and a chord is at most a diameter: 9 ≤ 2r, so r ≥ 4.5 cm.
Try one yourself
Can a circle of radius 3 cm pass through two points that are 7 cm apart? Explain.
Show answer
No. The least radius of a circle through two points 7 cm apart is 7 ÷ 2 = 3.5 cm, and 3 cm is smaller than that.
More questions like this
- 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? - Equal chords of a circle subtend equal angles at the centre of the circle.
- Chords of a circle that subtend equal angles at the centre are equal.
- Show that the triangle formed by a chord and the centre of the circle is isosceles.
- Show that if two such isosceles triangles (formed by a chord and the centre of the circle) have equal base length, they are congruent to each other.