Four questions to open the chapter with. Tap to bring up the next one.
Probe and ponder · 1
Three friends stand on a playground, no two of them in a line. Can you always draw one circle passing through where all three are standing?
Probe and ponder · 2
Two chords of a circle measure exactly the same length. Without measuring anything else — what must be true about the angle each makes at the centre?
Probe and ponder · 3
You stand at the centre of a circular ground looking at a bench on the boundary. Your friend stands on the boundary, looking at the same bench. Whose angle is bigger?
Probe and ponder · 4
Four points, no three in a line, all happen to sit on one circle. What does that force to be true about a pair of their angles?
Chapter 5 · What we will cover
Twelve theorems, one tool
5.2The circumcirclethrough 2 points, through 3 points
5.4Chords, bisectors & distanceequal chords · perpendicular bisector · how far from O
5.5Angles made by an arccentre vs. circumference · the semicircle
5.6–5.7Concyclic points & cyclic quadrilateralswhen four points share one circle
One big ideaएक बड़ा विचारEvery fact in this chapter comes from one tool: two radii of the same circle are always equal — which makes an isosceles triangle you can prove things with.
5.2 · Circles through two and three points
Building the Circumcircle
Through one point: infinite circles. Through two: still infinite — any centre on the perpendicular bisector works.
Step 1
Start with three non-collinear points, A, B, C.
Step 2
Draw the perpendicular bisector of AB, then of BC. They meet at exactly one point, O.
Step 3
O is equidistant from A, B and C — so a circle centred at O passes through all three.
Circumcircleपरिवृत्तA unique circle passes through three non-collinear points. Its centre O — the circumcentre — is where the perpendicular bisectors of the sides meet.
5.2 · Proof
Why exactly one circle?
1 · Equidistant from A, B
O sits on the perpendicular bisector of AB, so OA = OB.
2 · Equidistant from B, C
O sits on the perpendicular bisector of BC, so OB = OC.
3 · So OA = OB = OC
All three distances are equal to one radius, r.
4 · Why it's the ONLY circle
Any circle through A, B, C needs a centre on both bisectors — and two non-parallel lines meet at just one point.
Watch the trapसावधानThis 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.
Try it yourself
A triangular plot, one tank
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?
Spot the right angle
6² + 8² = 36 + 64 = 100 = 10² → the triangle is right-angled.
Use the shortcut
For a right triangle, the circumcentre is the midpoint of the hypotenuse — proved later as a case of the semicircle rule.
Answer
Circumradius = 10 ÷ 2 = 5 m from every corner.
Watch out
The trap
"Any three points determine a circle."
Why it fails
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.
5.4 · Chords and the angles they subtend
Equal Chords, Equal Angles
Two chords, AB and CD, in the same circle. Every radius is the same length — that's the entire engine here.
If AB = CD
Then ∠AOB = ∠COD — equal chords, equal central angles.
Converse, too
If ∠AOB = ∠COD to start, then AB = CD follows just as surely.
5.4 · Proof
Two triangles, one SSS
1 · Radii match
OA = OC and OB = OD — all radii of the same circle.
2 · Chords match
AB = CD — given.
3 · Triangles congruent
△AOB ≅ △COD by SSS.
4 · Angles match
So ∠AOB = ∠COD, by CPCT.
Converse (Thm 3)विलोमRun it backwards: given ∠AOB = ∠COD, the SAME triangles are congruent by SAS instead — so AB = CD by CPCT.
Try it yourself
Two 60° chords
Two chords of a circle of radius 7 cm each subtend 60° at the centre. Find the length of each chord.
Spot the shape
Each chord + two radii = an isosceles triangle with a 60° vertex angle between two equal sides — that's actually equilateral.
Answer
Each chord = the radius = 7 cm.
Watch out
The trap
"Equal chords in ANY two circles subtend equal angles."
Why it fails
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.
5.4 · Perpendicular from the centre
The Perpendicular Bisector of a Chord
Drop a line from centre O straight down to chord PQ, hitting it at exactly 90°. Where does it land?
Thm 4
The line from O to the midpoint of PQ is perpendicular to it.
Thm 5 (converse)
The perpendicular from O to PQ bisects it — lands exactly at the midpoint M.
5.4 · Proof
One RHS, run both ways
1 · Two right triangles
△OMP and △OMQ: OP = OQ (radii), OM common, ∠OMP = ∠OMQ = 90°.
2 · Congruent by RHS
△OMP ≅ △OMQ → PM = QM by CPCT.
3 · Backwards, too
Start from PM = QM with OM common: OP = OQ (radii) gives SSS instead, so ∠OMP = ∠OMQ = 90°.
Try it yourself
A 16 cm chord
A chord of length 16 cm sits in a circle of radius 10 cm. How far is it from the centre?
Only the perpendicular one does. Draw any other line from O to a random point on PQ — it will not split the chord evenly.
5.4 · Distance of a chord from the centre
Equal Chords, Equal Distance
Every chord's distance from O and its half-length obey one Pythagorean relation: d² + h² = R².
Thm 6 & 7
Equal chords are equidistant from O — and the converse holds too.
Thm 8
Of two unequal chords, the longer one always sits closer to the centre.
5.4 · Proof
One equation explains all three
1 · Halves from Thm 5
PM = ½PQ and SN = ½ST. If PQ = ST then PM = SN.
2 · RHS again
△OMP ≅ △ONS (OP = OS radii, PM = SN, right angles) → OM = ON.
3 · The Thm 8 shortcut
Since d² = R² − h², a bigger half-chord h forces a smaller distance d. Longer chord, closer to O — no new proof needed.
Try it yourself
24 cm vs. 10 cm
A circle has radius 13 cm. One chord is 24 cm, another is 10 cm. Which is closer to the centre?
Two quick Pythagoras checks
24 cm chord: 13² − 12² = 25 → d = 5 cm. 10 cm chord: 13² − 5² = 144 → d = 12 cm.
Answer
The 24 cm chord (5 cm away) is closer — confirming Theorem 8.
Watch out
The trap
"Equidistant just means they look close together on the page."
Why it fails
"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.
5.5 · Angles subtended by an arc
Centre Angle = 2 × Circumference Angle
Stand at the centre and look at arc BC. Then stand at a point A on the circle and look at the same arc.
The relationship
∠BOC is always exactly double ∠BAC — for any point A on the major arc.
The corollary
If BC is a diameter, ∠BOC = 180°, so ∠BAC = 90° — the angle in a semicircle.
5.5 · Proof
Two isosceles triangles, added up
1 · Join AO, extend to D
In △OAB: OA = OB (radii) → ∠OAB = ∠OBA.
2 · Exterior angle
∠BOD = ∠OAB + ∠OBA = 2∠OAB (exterior angle of a triangle).
3 · Same for the other side
Likewise, ∠COD = 2∠OAC.
4 · Add them
∠BOC = ∠BOD + ∠COD = 2(∠OAB + ∠OAC) = 2∠BAC.
Try it yourself
A diameter and 35°
PQ is a diameter. R is a point on the circle with ∠QPR = 35°. Find ∠PRQ and ∠PQR.
Use the corollary first
∠PRQ is the angle in a semicircle, so ∠PRQ = 90° straight away.
"The doubling rule works no matter which arc I pick."
Why it fails
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.
5.6 · Concyclic points
Equal Angles → Concyclic
Points C and D, on the same side of AB, both look at AB under the exact same angle.
The claim
If ∠ACB = ∠ADB, then A, B, C, D all sit on one circle.
The proof idea
Draw the circle through A, B, C (unique, by Thm 1). Show D can be neither inside nor outside it.
5.6 · Proof by contradiction
D has nowhere else to go
1 · Suppose D is inside
Extend CD to meet the circle at D′. Then ∠ADB, an exterior angle, is bigger than ∠AD′B = ∠ACB.
2 · That's a contradiction
We were given ∠ADB = ∠ACB. The same contradiction hits if D is outside instead.
3 · So D is ON the circle
Neither inside nor outside is possible — the only place left is on it.
Try it yourself
Two right angles on AB
C, D are on the same side of AB, with ∠ACB = ∠ADB = 90°. What can you conclude?
Combine two theorems
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.
Answer
A, B, C, D lie on a circle with AB as its diameter.
Watch out
The trap
"Equal angles at C and D always mean concyclic, whichever side they're on."
Why it fails
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.
5.7 · Cyclic quadrilaterals
Opposite Angles Sum to 180°
ABCD has all four corners on one circle. Add up one pair of opposite angles.
Thm 11
∠A + ∠C = 180°, and ∠B + ∠D = 180° too.
Thm 12 (converse)
If a quadrilateral's opposite angles sum to 180°, its vertices are concyclic.
5.7 · Proof
Theorem 9, used twice
1 · The reflex angle
∠BAD sees arc BCD from A: ∠BAD = ½ × reflex ∠BOD.
2 · The non-reflex angle
∠BCD sees arc BAD from C: ∠BCD = ½ × non-reflex ∠BOD.
3 · Add them
∠BAD + ∠BCD = ½ × 360° = 180°. The two arcs make the whole circle.
Converse (Thm 12)विलोमSame 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°.
Try it yourself
Solve for x
ABCD is cyclic, with ∠A = 3x° and ∠C = (x + 40)°. Find x, and both angles.
Set up the equation
3x + x + 40 = 180 → 4x = 140 → x = 35.
Answer
∠A = 105°, ∠C = 75°.
Watch out
The trap
"Opposite angles of ANY quadrilateral sum to 180°."
Why it fails
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.
Chapter 5 · Everything, at a glance
Four ideas, twelve theorems
1The circumcircle — three non-collinear points, one unique circle.
2–3Equal chords ↔ equal central angles — one SSS/SAS pair of triangles.
4–8Perpendicular bisector & distance from centre — all from d² + h² = R².
9Centre angle = 2 × circumference angle — and the 90° semicircle corollary.
10–12Concyclic points & cyclic quadrilaterals — same-side angles, and opposite angles summing to 180°.
Chapter 5 · Quick recap
Before you close the chapter
Q1Why do three points on a straight line have no circumcircle?
Q2State the relationship between the angle at the centre and at the circumference.
Q3What must be true for a quadrilateral to be cyclic?
Chapter Slides
Twelve theorems. One circle at a time.
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