Class 9 › Mathematics › Chapter
Circles: Chords, Angles and Cyclic Quadrilaterals
Definition and symmetry of a circle, circles through points, circumcircle, properties of chords, angle subtended by an arc, concyclic points and cyclic quadrilaterals, with the twelve theorems and their reasoning.
1. The Circle and its Parts
Circle: the set of all points in a plane that are equidistant from a fixed point (the centre). The common distance is the radius.
Locus: the set of all points satisfying a given condition. So a circle is the locus of points at a fixed distance from a given point. All figures here are drawn in a two-dimensional plane.
Locus: the set of all points satisfying a given condition. So a circle is the locus of points at a fixed distance from a given point. All figures here are drawn in a two-dimensional plane.
Centre, radius, chord and diameter.
| Term | Meaning |
|---|---|
| Radius | Distance from the centre to any point on the circle. All radii of a circle are equal. |
| Chord | Line segment joining two points on the circle. |
| Diameter | A chord through the centre; length $= 2 \times$ radius. It is the longest chord. |
| Arc | A connected portion of the circle between two points. The smaller is the minor arc, the larger the major arc (a diameter gives two semicircles). |
| Angle subtended at the centre | For a chord $BC$: $\angle BOC$. For an arc: the angle swept by the radius moving along the arc from one end to the other (less than $180^\circ$ for a minor arc, more than $180^\circ$ for a major arc). |
- The locus of points equidistant from two given points $A, B$ is the perpendicular bisector of $AB$ (every point on it is equidistant, and every equidistant point lies on it).
- A circle can be described by centre and radius; a line cuts a circle in at most two points.
2. Symmetries of a Circle
Rotational
Any angle
Rotating a circle about its centre by any angle leaves it unchanged (complete rotational symmetry). A rotating wheel looks the same at all times.
Reflection
Every diameter
Every diameter is a line of reflection symmetry, so there are infinitely many lines of symmetry; each passes through the centre.
Finding the centre
Folding
Two folds along diameters meet at the centre. Alternatively, the perpendicular bisectors of two chords meet at the centre.
- Rotating a circle carries a chord to another chord of the same length; this is the idea behind the equal-chord theorems.
- For comparison, a regular $n$-gon has only $n$ rotational symmetries (rotations by multiples of $360^\circ/n$) and $n$ lines of reflection.
3. How Many Circles Pass Through Given Points?
One point
Infinitely many
Any point can be the centre of a circle through it.
Two points $A, B$
Infinitely many
The centre must satisfy $OA = OB$, so it lies on the perpendicular bisector of $AB$. Every point of that line is the centre of a circle through $A$ and $B$.
Three collinear points
None
The perpendicular bisectors of $AB$ and $BC$ are parallel and never meet, so no point is equidistant from all three. A line cannot meet a circle in three points.
Three non-collinear points
Exactly one
Theorem 1 below.
- Smallest circle through $A, B$: the one with centre at the midpoint of $AB$ and radius $\tfrac12 AB$ ($AB$ is a diameter). So the radius is at least $\tfrac12 AB$; there is no largest circle.
- As the centre moves farther from $AB$ along the perpendicular bisector, the radius increases and the circle looks less curved (closer to a straight line near $AB$).
- Radius of any circle through $A, B$ with centre at distance $d$ from the midpoint: $r = \sqrt{(\tfrac12 AB)^2 + d^2}$.
4. Circle Through Three Points: the Circumcircle
Theorem 1
There is a unique circle passing through three non-collinear points.
Reasoning
1. If a circle passes through $A, B, C$ with centre $O$, then $OA = OB$, so $O$ is on the perpendicular bisector of $AB$.
2. Also $OA = OC$, so $O$ is on the perpendicular bisector of $AC$.
3. As $A, B, C$ are not collinear, these two bisectors are not parallel and meet at exactly one point; that point must be $O$.
4. The circle with centre $O$ and radius $OA$ passes through $A, B, C$, and there is only one such circle.
Circumcircle of $\triangle ABC$: the circle through its three vertices. Its centre is the circumcentre (intersection of the perpendicular bisectors of the sides) and its radius the circumradius. The triangle is inscribed in the circle and the circle circumscribes the triangle.
| Triangle | Position of circumcentre |
|---|---|
| Acute-angled | Inside the triangle |
| Obtuse-angled | Outside the triangle |
| Right-angled | Midpoint of the hypotenuse |
- All three perpendicular bisectors of the sides meet at the circumcentre (they are concurrent).
- To draw the circumcircle: draw perpendicular bisectors of two sides, mark their intersection $O$, and draw a circle with radius $OA$.
- Congruent triangles can share the same circumcircle (same circle, different positions).
5. Chords and the Angles They Subtend at the Centre
Theorem 2 Equal chords of a circle subtend equal angles at the centre.
Theorem 3 (converse) Chords that subtend equal angles at the centre are equal.
Theorem 3 (converse) Chords that subtend equal angles at the centre are equal.
Proof of Theorem 2
Given: chords $AB = DE$ in a circle with centre $C$. To show: $\angle ACB = \angle DCE$.
$CA = CD$ and $CB = CE$ (radii) and $AB = DE$ (given). By SSS, $\triangle CAB \cong \triangle CDE$, so $\angle ACB = \angle DCE$ (corresponding parts).
Proof of Theorem 3
Given: $\angle ACB = \angle DCE$. To show: $AB = DE$.
$CA = CD$, $CB = CE$ (radii) and the included angles are equal. By SAS, $\triangle ACB \cong \triangle DCE$, so $AB = DE$.
- The triangle formed by a chord and the two radii to its ends is isosceles (two equal radii); two such triangles with equal bases are congruent (SSS).
- Longer chord $\iff$ larger angle at the centre (for angles up to $180^\circ$).
6. Midpoint of a Chord and Perpendicular from the Centre
Theorem 4 The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.
Theorem 5 (converse) The perpendicular from the centre to a chord bisects the chord.
Theorem 5 (converse) The perpendicular from the centre to a chord bisects the chord.
$M$ is the midpoint of chord $AB$; $CM \perp AB$ and $d = CM$ is the distance from the centre to the chord.
Proof of Theorem 4
1. $CA = CB$ (radii) and $AM = MB$ ($M$ is the midpoint); $CM$ is common.
2. By SSS/SAS, $\triangle CMA \cong \triangle CMB$, so $\angle CMA = \angle CMB$.
3. They form a linear pair: $\angle CMA + \angle CMB = 180^\circ$, so each is $90^\circ$. Hence $CM \perp AB$.
Proof of Theorem 5 (converse)
1. Given $CM \perp AB$: $\angle CMA = \angle CMB = 90^\circ$, $CA = CB$ and $CM$ common.
2. By RHS, $\triangle CMA \cong \triangle CMB$, so $AM = BM$.
Chord length from radius and distance
$$\text{chord} = 2\sqrt{r^2 - d^2} \qquad r^2 = d^2 + \left(\tfrac{\text{chord}}{2}\right)^2$$
- Derived from the right triangle $CMA$ (Baudhāyana–Pythagoras Theorem): $CA^2 = CM^2 + AM^2$.
- Perpendicular bisector of a chord passes through the centre (since the centre lies on the bisector of every chord).
7. Distance of Chords from the Centre
"Distance of a chord from the centre" always means the perpendicular distance; by Theorem 5 this is $CM$ to the midpoint $M$.
Theorem 6 Chords of equal length are equidistant from the centre.
Theorem 7 (converse) Chords equidistant from the centre are equal in length.
Theorem 8 Of two unequal chords, the longer chord is nearer to the centre.
Theorem 7 (converse) Chords equidistant from the centre are equal in length.
Theorem 8 Of two unequal chords, the longer chord is nearer to the centre.
Proof of Theorem 6 (two ways)
Method 1: $\triangle CAB \cong \triangle CFG$ by SSS ($CA = CF$, $CB = CG$ radii; $AB = FG$). Corresponding altitudes of congruent triangles are equal, so $CE = CH$.
Method 2: $E, H$ are midpoints, so $AE = FH$; also $\angle CEA = \angle CHF = 90^\circ$ and $CA = CF$. By RHS, $\triangle CEA \cong \triangle CHF$, so $CE = CH$.
Method 3 (Pythagoras): $CE^2 = r^2 - (\tfrac12 AB)^2$ and $CH^2 = r^2 - (\tfrac12 FG)^2$; equal chords give equal distances.
Proof of Theorem 7
$CE = CH$ (given), $CA = CF$ (radii), right angles at $E$ and $H$. By RHS, $\triangle CEA \cong \triangle CHF$, so $AE = FH$; doubling (midpoints), $AB = FG$. (Or: $AB = 2\sqrt{r^2 - CE^2} = 2\sqrt{r^2 - CH^2} = FG$.)
Proof of Theorem 8
Given: $AB \gt DE$; $CF, CG$ are the perpendiculars to $AB, DE$. To show: $CF \lt CG$.
1. By Pythagoras: $CF^2 + AF^2 = AC^2 = r^2$ and $CG^2 + GD^2 = CD^2 = r^2$, so $CF^2 + AF^2 = CG^2 + GD^2$.
2. $F, G$ are midpoints, so $AF = \tfrac12 AB \gt \tfrac12 DE = GD$, giving $AF^2 \gt GD^2$.
3. Therefore $CF^2 \lt CG^2$, so $CF \lt CG$.
- Diameter: distance from centre $= 0$, the nearest and the longest chord. As a chord moves away, it shortens; when it shrinks to a point, its distance equals the radius.
- Distance and length move in opposite directions: the greater the length, the smaller the distance.
- Midpoints of all chords of a fixed length lie at the same distance $d$ from the centre, so they form a concentric circle of radius $d$.
8. Angle Subtended by an Arc
- At the centre: measure of $\angle AOB$ as the radius sweeps from $OA$ to $OB$ along the arc (minor arc: less than $180^\circ$; major arc: more than $180^\circ$, a reflex angle).
- At a point on the circle outside the arc: $\angle ACB$ where $C$ lies on the circle but not on that arc. Remarkably, it does not depend on where $C$ is on the remaining part of the circle.
Theorem 9
The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the circle outside the arc.
$$\angle AOB = 2\,\angle ADB$$
The arc $AB$ subtends $2\theta$ at the centre $O$ and the same angle $\theta$ at every point ($D$, $D'$, ...) of the major arc.
Proof (centre $C$, arc $AFB$, point $D$ on the circle outside the arc)
Case 1: $DC$ extended meets the circle at $E$ on the arc $AFB$.
1. $CB = CD$ (radii), so $\triangle DCB$ is isosceles: $\angle CBD = \angle CDB$.
2. $\angle BCE$ is an exterior angle of $\triangle BCD$: $\angle BCE = \angle CBD + \angle CDB = 2\angle BDC$. Similarly $\angle ACE = 2\angle CDA$.
3. Add: $\angle BCA = \angle BCE + \angle ECA = 2(\angle BDC + \angle CDA) = 2\angle BDA$.
Case 2: $DC$ extended meets the circle at $E$ outside the arc.
Same isosceles/exterior-angle step: $\angle ACE = 2\angle ADC$ and $\angle BCE = 2\angle BDC$. Subtract: $\angle ACB = \angle ACE - \angle BCE = 2(\angle ADC - \angle BDC) = 2\angle ADB$.
Consequence
Angles in the same segment
All angles subtended by an arc at points on the remaining part of the circle are equal (each is half the central angle).
Corollary
Angle in a semicircle
The angle subtended by a diameter at any point on the circle is $90^\circ$ (central angle is $180^\circ$, half is $90^\circ$).
Warning
Only for points on the circle
For points inside or outside the circle, the angle subtended by the chord differs from the circle value (it is larger inside, smaller outside).
Special case
Central angle $60^\circ$
The chord equals the radius (equilateral triangle $OAB$); angle at the circle is $30^\circ$.
Corollary means a fact that follows immediately from an already proved result. Equal angles at the circle imply equal chords and equal arcs (via the central angle).
9. Concyclic Points
Concyclic points: points that all lie on the same circle. Three non-collinear points are always concyclic (Theorem 1); four points need a condition.
Theorem 10
If a segment $AB$ subtends equal angles at two other points $C, D$ lying on the same side of $AB$, then $A, B, C, D$ are concyclic.
- This is the converse of "angles in the same segment are equal".
Reasoning (proof by contradiction)
1. $A, B, C$ are non-collinear, so there is a circle through them (Theorem 1). Suppose $D$ is not on it: $D$ is outside or inside.
2. Join $AD$ and let it (extended if needed) meet the circle at $E$. Since $C$ and $E$ are on the same side of $AB$ on the circle, $\angle ACB = \angle AEB$.
3. If $D$ is outside, $\angle AEB$ is an exterior angle of $\triangle BED$, so $\angle AEB \gt \angle ADB$. But $\angle AEB = \angle ACB = \angle ADB$, a contradiction. Inside is similar.
4. So $D$ lies on the circle: $A, B, C, D$ are concyclic.
- Test for concyclicity: equal angles on the same side of a segment, or (for a quadrilateral) opposite angles summing to $180^\circ$ (Theorem 12).
- Equal angles at points on opposite sides of $AB$ do not make the points concyclic.
10. Cyclic Quadrilaterals
Cyclic quadrilateral: a quadrilateral (4-gon) whose four vertices are concyclic (it is inscribed in a circle).
Theorem 11 The sum of each pair of opposite angles of a cyclic quadrilateral is $180^\circ$.
Theorem 12 (converse) If a quadrilateral has opposite angles that add up to $180^\circ$, it is cyclic.
Theorem 12 (converse) If a quadrilateral has opposite angles that add up to $180^\circ$, it is cyclic.
A cyclic quadrilateral $ABCD$: opposite angles are supplementary.
Proof of Theorem 11
1. $A$ lies on the circle outside arc $BCD$, so $\angle BAD = \tfrac12$ (angle subtended at $O$ by arc $BCD$), i.e. half the reflex angle $BOD$.
2. $C$ lies outside arc $BAD$, so $\angle BCD = \tfrac12 \angle BOD$ (the non-reflex angle at $O$).
3. Add: $\angle BAD + \angle BCD = \tfrac12$ (complete angle at $O$) $= \tfrac12 \times 360^\circ = 180^\circ$. The same holds for $\angle B + \angle D$.
Reasoning for Theorem 12 (by contradiction)
1. Given $\angle A + \angle C = 180^\circ$. Draw the circle through $A, B, D$. Suppose $C$ is not on it; say $CD$ meets the circle at $E$.
2. $ABED$ is cyclic, so $\angle BAD + \angle BED = 180^\circ$ (Theorem 11). With the given condition, $\angle BCD = \angle BED$.
3. But $\angle BED$ is an exterior angle of $\triangle BEC$, so $\angle BED \gt \angle BCD$, a contradiction. (The case with $C$ inside is similar.) Hence $C$ is on the circle.
- Exterior angle property: the exterior angle at a vertex of a cyclic quadrilateral equals the interior opposite angle (since the interior angle and its opposite add to $180^\circ$ and the exterior angle is supplementary to the interior one).
- Checking whether a quadrilateral is cyclic: test whether a pair of opposite angles sums to $180^\circ$ (if one pair does, the other automatically does too).
- If $A$ and $C$ are opposite angles and $A + C \neq 180^\circ$, no circle passes through all four vertices.
11. Useful Results and Applications
Right triangle
Hypotenuse is a diameter
The circumcentre of a right triangle is the midpoint of the hypotenuse, so circumradius $= \tfrac12$ hypotenuse; conversely, an angle inscribed in a semicircle is $90^\circ$.
Isosceles triangle
Altitude through centre
In an inscribed isosceles triangle $ABC$ ($AB = AC$) the altitude from $A$ passes through the centre (it is the perpendicular bisector of $BC$).
Equal chords
Angle bisector through centre
If chords $AB$ and $AC$ are equal, the centre lies on the bisector of $\angle BAC$ (the centre is equidistant from both chords).
Parallel chords
Distance between them
With distances $d_1$ and $d_2$ from the centre: the chords are $d_1 + d_2$ apart if on opposite sides of the centre and $|d_1 - d_2|$ apart if on the same side (find each $d$ from $d = \sqrt{r^2 - (\text{chord}/2)^2}$).
Shortest chord
Through an interior point $A$
Chord length $= 2\sqrt{r^2 - d^2}$ with $d \le OA$; it is least when $d = OA$, i.e. when the chord is perpendicular to $OA$.
Quadrilaterals
Parallelogram / rectangle
A cyclic parallelogram has equal opposite angles that add to $180^\circ$, so each is $90^\circ$: a rectangle is the only parallelogram that can be inscribed in a circle. Its diagonals are diameters, so they meet at the centre.
Regular hexagon
Inscribed hexagon
Each side subtends $60^\circ$ at the centre, so side $= r$; distance of each side from the centre $= \dfrac{\sqrt{3}}{2}r$.
Drawing
Circle with given chord and distance
A chord of length $2m$ at distance $d$ has radius $\sqrt{m^2 + d^2}$; the circle is the circumcircle of a suitable triangle.
Diameter
Longest chord
No chord exceeds the diameter: for a chord at distance $d \ge 0$, length $= 2\sqrt{r^2 - d^2} \le 2r$.
How geometric proofs are written
Given (hypothesis) and To show (conclusion), a clear figure, then reasons using definitions, earlier theorems and congruence criteria (SSS, SAS, ASA, RHS), ending with the conclusion.
A converse swaps the "given" and "to show" parts; a converse is a new statement and needs its own proof. A few examples do not prove a statement.
12. Theorem Sheet and Quick Revision
| No. | Statement | Key reason |
|---|---|---|
| 1 | Unique circle through three non-collinear points | Perpendicular bisectors of two sides meet at one point |
| 2 | Equal chords subtend equal angles at the centre | SSS congruence |
| 3 | Chords subtending equal angles at the centre are equal | SAS congruence |
| 4 | Centre to midpoint of chord is perpendicular to the chord | Isosceles triangle / congruence, linear pair |
| 5 | Perpendicular from centre bisects the chord | RHS congruence |
| 6 | Equal chords are equidistant from the centre | SSS or RHS congruence; Pythagoras |
| 7 | Chords equidistant from the centre are equal | RHS congruence |
| 8 | Longer chord is closer to the centre | Pythagoras: $CF^2 + AF^2 = CG^2 + GD^2$ |
| 9 | Arc subtends twice the angle at the centre as at the circle | Isosceles triangles + exterior angle |
| Cor. | Angle in a semicircle is $90^\circ$ | Theorem 9 with central angle $180^\circ$ |
| 10 | Equal angles on the same side of a segment imply concyclic | Contradiction using Theorem 1 |
| 11 | Opposite angles of a cyclic quadrilateral are supplementary | Theorem 9 and full angle $360^\circ$ |
| 12 | Opposite angles supplementary implies cyclic | Contradiction using Theorem 11 |
Formulas
$$\text{chord} = 2\sqrt{r^2 - d^2}, \qquad \angle\text{at centre} = 2 \times \angle\text{at circle}, \qquad \angle A + \angle C = 180^\circ \;(\text{cyclic})$$
Common mistakes:
• Using the full chord length instead of half the chord in the right triangle with the radius.
• Taking the distance of a chord from the centre along a slanted line instead of the perpendicular.
• Applying Theorem 9 to a point on the arc instead of on the remaining part of the circle.
• Forgetting the reflex angle when the arc is a major arc.
• Assuming any quadrilateral is cyclic; it must satisfy the opposite-angle condition.
• Mixing up Theorems and their converses (they need separate proofs).
• Using the full chord length instead of half the chord in the right triangle with the radius.
• Taking the distance of a chord from the centre along a slanted line instead of the perpendicular.
• Applying Theorem 9 to a point on the arc instead of on the remaining part of the circle.
• Forgetting the reflex angle when the arc is a major arc.
• Assuming any quadrilateral is cyclic; it must satisfy the opposite-angle condition.
• Mixing up Theorems and their converses (they need separate proofs).