Through one point: infinite circles. Through two: still infinite — any centre on the perpendicular bisector works.
Start with three non-collinear points, A, B, C.
Draw the perpendicular bisector of AB, then of BC. They meet at exactly one point, O.
O is equidistant from A, B and C — so a circle centred at O passes through all three.
O sits on the perpendicular bisector of AB, so OA = OB.
O sits on the perpendicular bisector of BC, so OB = OC.
All three distances are equal to one radius, r.
Any circle through A, B, C needs a centre on both bisectors — and two non-parallel lines meet at just one point.
A triangular plot has sides 6 m, 8 m and 10 m. A tank must sit equidistant from all three corners. How far from each corner?
6² + 8² = 36 + 64 = 100 = 10² → the triangle is right-angled.
For a right triangle, the circumcentre is the midpoint of the hypotenuse — proved later as a case of the semicircle rule.
Circumradius = 10 ÷ 2 = 5 m from every corner.
"Any three points determine a circle."
Only true when the points are non-collinear. Three points on a line have no circumcircle at all — their perpendicular bisectors are parallel and never meet.
Two chords, AB and CD, in the same circle. Every radius is the same length — that's the entire engine here.
Then ∠AOB = ∠COD — equal chords, equal central angles.
If ∠AOB = ∠COD to start, then AB = CD follows just as surely.
OA = OC and OB = OD — all radii of the same circle.
AB = CD — given.
△AOB ≅ △COD by SSS.
So ∠AOB = ∠COD, by CPCT.
Two chords of a circle of radius 7 cm each subtend 60° at the centre. Find the length of each chord.
Each chord + two radii = an isosceles triangle with a 60° vertex angle between two equal sides — that's actually equilateral.
Each chord = the radius = 7 cm.
"Equal chords in ANY two circles subtend equal angles."
The theorem needs the same circle (or congruent circles). A 5 cm chord in a small circle sweeps a much bigger central angle than a 5 cm chord in a huge one.
Drop a line from centre O straight down to chord PQ, hitting it at exactly 90°. Where does it land?
The line from O to the midpoint of PQ is perpendicular to it.
The perpendicular from O to PQ bisects it — lands exactly at the midpoint M.
△OMP and △OMQ: OP = OQ (radii), OM common, ∠OMP = ∠OMQ = 90°.
△OMP ≅ △OMQ → PM = QM by CPCT.
Start from PM = QM with OM common: OP = OQ (radii) gives SSS instead, so ∠OMP = ∠OMQ = 90°.
A chord of length 16 cm sits in a circle of radius 10 cm. How far is it from the centre?
Half-chord = 8 cm. distance² = 10² − 8² = 100 − 64 = 36.
Distance from centre = 6 cm.
"Any line from the centre to a chord bisects it."
Only the perpendicular one does. Draw any other line from O to a random point on PQ — it will not split the chord evenly.
Every chord's distance from O and its half-length obey one Pythagorean relation: d² + h² = R².
Equal chords are equidistant from O — and the converse holds too.
Of two unequal chords, the longer one always sits closer to the centre.
PM = ½PQ and SN = ½ST. If PQ = ST then PM = SN.
△OMP ≅ △ONS (OP = OS radii, PM = SN, right angles) → OM = ON.
Since d² = R² − h², a bigger half-chord h forces a smaller distance d. Longer chord, closer to O — no new proof needed.
A circle has radius 13 cm. One chord is 24 cm, another is 10 cm. Which is closer to the centre?
24 cm chord: 13² − 12² = 25 → d = 5 cm.
10 cm chord: 13² − 5² = 144 → d = 12 cm.
The 24 cm chord (5 cm away) is closer — confirming Theorem 8.
"Equidistant just means they look close together on the page."
"Distance from the centre" is one specific measured length — the perpendicular from O to the chord. Two chords can look close in a sketch with very different real distances.
Stand at the centre and look at arc BC. Then stand at a point A on the circle and look at the same arc.
∠BOC is always exactly double ∠BAC — for any point A on the major arc.
If BC is a diameter, ∠BOC = 180°, so ∠BAC = 90° — the angle in a semicircle.
In △OAB: OA = OB (radii) → ∠OAB = ∠OBA.
∠BOD = ∠OAB + ∠OBA = 2∠OAB (exterior angle of a triangle).
Likewise, ∠COD = 2∠OAC.
∠BOC = ∠BOD + ∠COD = 2(∠OAB + ∠OAC) = 2∠BAC.
PQ is a diameter. R is a point on the circle with ∠QPR = 35°. Find ∠PRQ and ∠PQR.
∠PRQ is the angle in a semicircle, so ∠PRQ = 90° straight away.
∠PRQ = 90°. Angle sum: ∠PQR = 180 − 90 − 35 = 55°.
"The doubling rule works no matter which arc I pick."
You must compare the centre angle to a point on the arc it does NOT enclose. Pick the wrong arc and you're really looking at the reflex angle at O instead — a very common exam slip.
Points C and D, on the same side of AB, both look at AB under the exact same angle.
If ∠ACB = ∠ADB, then A, B, C, D all sit on one circle.
Draw the circle through A, B, C (unique, by Thm 1). Show D can be neither inside nor outside it.
Extend CD to meet the circle at D′. Then ∠ADB, an exterior angle, is bigger than ∠AD′B = ∠ACB.
We were given ∠ADB = ∠ACB. The same contradiction hits if D is outside instead.
Neither inside nor outside is possible — the only place left is on it.
C, D are on the same side of AB, with ∠ACB = ∠ADB = 90°. What can you conclude?
By this theorem, A, B, C, D are concyclic. Since 90° is the semicircle angle, AB itself must be the diameter — Theorem 9's corollary, used in reverse.
A, B, C, D lie on a circle with AB as its diameter.
"Equal angles at C and D always mean concyclic, whichever side they're on."
The "same side" condition is load-bearing. Equal angles on OPPOSITE sides is a different situation entirely — that's the cyclic-quadrilateral case, coming up next.
ABCD has all four corners on one circle. Add up one pair of opposite angles.
∠A + ∠C = 180°, and ∠B + ∠D = 180° too.
If a quadrilateral's opposite angles sum to 180°, its vertices are concyclic.
∠BAD sees arc BCD from A: ∠BAD = ½ × reflex ∠BOD.
∠BCD sees arc BAD from C: ∠BCD = ½ × non-reflex ∠BOD.
∠BAD + ∠BCD = ½ × 360° = 180°. The two arcs make the whole circle.
ABCD is cyclic, with ∠A = 3x° and ∠C = (x + 40)°. Find x, and both angles.
3x + x + 40 = 180 → 4x = 140 → x = 35.
∠A = 105°, ∠C = 75°.
"Opposite angles of ANY quadrilateral sum to 180°."
Only true when it's cyclic. A generic quadrilateral's four angles total 360°, but the two opposite pairs can split that any way at all — 180/180 is special.
The circumcircle — three non-collinear points, one unique circle.
Equal chords ↔ equal central angles — one SSS/SAS pair of triangles.
Perpendicular bisector & distance from centre — all from d² + h² = R².
Centre angle = 2 × circumference angle — and the 90° semicircle corollary.
Concyclic points & cyclic quadrilaterals — same-side angles, and opposite angles summing to 180°.
A unique circle passes through three non-collinear points. Its centre O — the circumcentre — is where the perpendicular bisectors of the sides meet.
This only works when A, B, C are non-collinear. Three points on one straight line have PARALLEL perpendicular bisectors — they never meet, so no circle passes through all three.
Run it backwards: given ∠AOB = ∠COD, the SAME triangles are congruent by SAS instead — so AB = CD by CPCT.
Same contradiction argument as Theorem 10: draw the circle through A, B, C — if D isn't on it, the angle sum can never land exactly on 180°.
Why do three points on a straight line have no circumcircle?
State the relationship between the angle at the centre and at the circumference.
What must be true for a quadrilateral to be cyclic?