Q1.
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?
Answer
There is no circle at all through three collinear points.
Given: A, B, C on one line, all different. To show: no point O has OA = OB = OC.
Suppose such a circle existed, with centre O.
OA = OB ⇒ O lies on the perpendicular bisector of AB
OB = OC ⇒ O lies on the perpendicular bisector of BC
But AB and BC lie along the same line ℓ.
So both perpendicular bisectors are perpendicular to ℓ ⇒ they are parallel to each other.
They pass through different points (the midpoints of AB and of BC are different), so they are distinct parallel lines — they never meet.
Hence no such point O exists. No circle passes through three collinear points.
OA = OB ⇒ O lies on the perpendicular bisector of AB
OB = OC ⇒ O lies on the perpendicular bisector of BC
But AB and BC lie along the same line ℓ.
So both perpendicular bisectors are perpendicular to ℓ ⇒ they are parallel to each other.
They pass through different points (the midpoints of AB and of BC are different), so they are distinct parallel lines — they never meet.
Hence no such point O exists. No circle passes through three collinear points.
Why it happens: Theorem 1 works because two perpendicular bisectors of a genuine triangle are not parallel, so they cross at exactly one point. Collinearity destroys precisely that step: the two bisectors become parallel, the meeting point vanishes, and with it the centre.
Check it yourself: the same fact read the other way round says a straight line can meet a circle in at most 2 points — never 3.