Chapter 5: I'm Up and Down, and Round and Round (Circles) - Ultimate 100/100 Master Study Guide

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Topic 1: Circle Definitions, Locus & Key Terminology

Core Definitions & Locus Concept

Circle: The locus (collection of all points) in a 2D plane that are at a fixed constant distance from a fixed point.
Centre ($O$): The fixed point from which all boundary points are equidistant.
Radius ($r$): The fixed distance segment from centre $O$ to any boundary point $P$.
Chord ($BC$): A straight line segment joining any two points $B$ and $C$ on the circle boundary.
Diameter ($d$): A chord that passes directly through the centre $O$. It is the longest chord of a circle ($d = 2r$).
Secant: A line that intersects a circle at two distinct points.
Segment: The region bounded by a chord and the corresponding arc (Minor Segment & Major Segment).
Sector: The region bounded by two radii and the enclosed arc (Minor Sector & Major Sector).

Fig 5.1: Mathematical Components of a Circle (Centre O, Radius OA, Diameter DE, Chord BC)

Teacher's Exam Secret — Frequently Asked 1-Mark Concept

Remember: A diameter is the longest chord of a circle. All diameters are chords, but NOT all chords are diameters. Diameter has length $2r$ and passes through centre $O$.

Immediate Solved Exam Applications

Example 1.1: Longest Chord Calculation NCERT

Question: Find the length of the longest chord in a circle of radius $r = 7.5\text{ cm}$.

Solution:

The longest chord of a circle is its diameter ($d = 2r$).

$$\text{Diameter } d = 2 \times 7.5\text{ cm} = \mathbf{15\text{ cm}}$$

Topic 1: School Exam Practice Problem Kit

Problem 1.1 (Concentric Circles Geometry) 2 MARKS

Two concentric circles have radii $13\text{ cm}$ and $5\text{ cm}$. Find the length of the chord of the outer circle which touches the inner circle.

Step-by-Step Solution:

In right $\Delta OMA$, $OA = 13\text{ cm}$, $OM = 5\text{ cm}$. $$AM = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$$ $$\text{Total Chord Length } AB = 2 \times AM = 2 \times 12\text{ cm} = \mathbf{24\text{ cm}}$$

Fig 5.1A: Concentric Circles Chord $AB$

Problem 1.2 (Locus Reasoning) 1 MARK

Describe the locus of the center of a circle of radius $5\text{ cm}$ which rolls along the outside of a fixed circle of radius $8\text{ cm}$.

Step-by-Step Solution:

The locus is a concentric circle with radius $= R + r = 8 + 5 = \mathbf{13\text{ cm}}$.

Topic 2: Symmetries of a Circle & Number of Circles Through Points

Symmetry & Point Rules

Rotational Symmetry: Infinite / continuous rotational symmetry around centre $O$ through any angle $\theta$.
Reflection Symmetry: Every line containing a diameter is a line of reflection symmetry (infinitely many lines of symmetry).
Circles Through 1 Point: Infinitely many circles can pass through a single point $A$.
Circles Through 2 Points ($A, B$): Infinitely many circles can pass through $A$ and $B$. All their centres lie on the perpendicular bisector of line segment $AB$.
Smallest Circle: The circle with $AB$ as diameter (radius $= \frac{1}{2}AB$).
Circles Through 3 Collinear Points: ZERO circles! (Because perpendicular bisectors of collinear segments are parallel and never intersect).

Theorem 1: Unique Circle Through 3 Non-Collinear Points

Statement: There is one and only one circle passing through three non-collinear points $A, B, C$.

Fig 5.3: Proof Diagram for Theorem 1 — Perpendicular Bisectors $L_1$ and $L_2$ Intersect at Unique Point $O$

Given: Three non-collinear points $A, B, C$.

To Prove: A unique circle passes through $A, B, C$.

Proof:

Join $AB$ and $BC$. Draw perpendicular bisector $L_1$ of $AB$ and $L_2$ of $BC$.
Since $A, B, C$ are non-collinear, lines $AB$ and $BC$ are not parallel. Hence their perpendicular bisectors $L_1$ and $L_2$ must intersect at a unique point, say $O$.
Since $O$ lies on $L_1$ (bisector of $AB$), $OA = OB$.
Since $O$ lies on $L_2$ (bisector of $BC$), $OB = OC$.
Therefore, $OA = OB = OC = R$. Taking $O$ as centre and radius $R = OA$, the circle passes through $A, B, C$. Since two non-parallel lines intersect at exactly ONE point, $O$ is unique. Thus, the circle is UNIQUE. ■

Positions of Circumcentre $O$ for Different Triangles

Type of Triangle $\Delta ABC$	Position of Circumcentre $O$	Reference
Acute-Angled Triangle ($\angle < 90^\circ$)	Lies INSIDE the triangle	Fig 5.4 (Acute Circumcircle)
Obtuse-Angled Triangle ($\angle > 90^\circ$)	Lies OUTSIDE the triangle	Fig 5.5 (Obtuse Circumcircle)
Right-Angled Triangle ($\angle = 90^\circ$)	Lies exactly at the MIDPOINT of Hypotenuse	Fig 5.6 (Right Triangle Circumcircle)

Topic 2: School Exam Practice Problem Kit

Problem 2.1 (Circumradius of Right Triangle) 2 MARKS

Find the radius of the circumcircle of a right-angled triangle with legs $12\text{ cm}$ and $5\text{ cm}$.

Step-by-Step Solution:

By Pythagoras Theorem, $\text{Hypotenuse } c = \sqrt{12^2 + 5^2} = 13\text{ cm}$. $$\text{Circumradius } R = \frac{\text{Hypotenuse}}{2} = \frac{13}{2} = \mathbf{6.5\text{ cm}}$$

Problem 2.2 (Geometric Proof) 3 MARKS

Prove that the perpendicular bisector of any chord of a circle must pass through the centre of the circle.

Step-by-Step Solution:

Let $AB$ be a chord and $M$ its midpoint. $OA = OB = r \implies \Delta OAB$ is isosceles. In an isosceles triangle, median $OM$ is perpendicular to base $AB \implies OM \perp AB$. Thus, perpendicular bisector passes through $O$.

Fig 5.2A: Chord Bisector $OM \perp AB$

Topic 3: Chords & Angles Subtended at the Centre

Theorem 2: Equal Chords Subtend Equal Angles at Centre

Statement: Equal chords of a circle subtend equal angles at the centre.

Fig 5.7: Proof Diagram for Theorem 2 — Equal Chords $AB = DE \iff \angle ACB = \angle DCE$

Given: A circle with centre $C$. Chords $AB = DE$.

To Prove: $\angle ACB = \angle DCE$.

Proof:

In $\Delta CAB$ and $\Delta CDE$:

$CA = CD = r$ (Radii of same circle)
$CB = CE = r$ (Radii of same circle)
$AB = DE$ (Given equal chords)

By SSS Congruence Rule, $\Delta CAB \cong \Delta CDE$.

$$\implies \mathbf{\angle ACB = \angle DCE} \quad (\text{by CPCTC})$$

■

Topic 3: School Exam Practice Problem Kit

Problem 3.1 (Chord Equal to Radius) NCERT EXERCISE

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at the centre.

Step-by-Step Solution:

In $\Delta OAB$, $OA = OB = AB = r$. Thus $\Delta OAB$ is equilateral $\implies \mathbf{\angle AOB = 60^\circ}$.

Fig 5.3A: Equilateral $\Delta OAB$

Topic 4: Midpoints & Perpendicular Bisectors of Chords

Theorem 4: Line Joining Centre to Midpoint is Perpendicular

Statement: The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Fig 5.8: Proof Diagram for Theorem 4 — Perpendicular $CM \perp AB \iff AM = BM$

Given: $M$ is midpoint of chord $AB$ ($AM = BM$) in circle with centre $C$.

To Prove: $CM \perp AB$ ($\angle CMA = \angle CMB = 90^\circ$).

Proof: In $\Delta CMA$ and $\Delta CMB$, $CA = CB = r$, $CM = CM$, $AM = BM$. By SSS congruence, $\Delta CMA \cong \Delta CMB \implies \angle CMA = \angle CMB = 90^\circ \implies \mathbf{CM \perp AB}$. ■

Topic 4: School Exam Practice Problem Kit

Problem 4.1 (Concentric Circles Secant Intersecting) 3 MARKS / NCERT

A straight line intersects two concentric circles with common centre $O$ at points $A, B, C, D$. Prove that $AB = CD$.

Step-by-Step Solution:

Draw $OM \perp AD$.
For outer circle chord $AD$: $AM = MD$.
For inner circle chord $BC$: $BM = MC$.
Subtracting: $AM - BM = MD - MC \implies \mathbf{AB = CD}$. ■

Fig 5.4A: Secant Line $AD$

Topic 5: Distance of Chords & Comparison of Unequal Chords

Theorem 6: Equal Chords are Equidistant from Centre

Statement: Chords of a circle having equal length are at equal distance from the centre.

Fig 5.9: Proof Diagram for Theorem 6 — Equal Chords $AB = FG \iff CE = CH$

Proof: Draw $CE \perp AB$ and $CH \perp FG$. By Theorem 5, $AE = AB/2$ and $FH = FG/2$. Since $AB = FG \implies AE = FH$. Right $\Delta CEA \cong \text{Right } \Delta CHF$ by RHS $\implies \mathbf{CE = CH}$. ■

Topic 5: School Exam Practice Problem Kit

Problem 5.1 (Intersecting Equal Chords) 5 MARKS / NCERT EXERCISE

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Step-by-Step Solution:

Draw $OL \perp AB$, $OM \perp CD$. Right $\Delta OLP \cong \text{Right } \Delta OMP$ by RHS $\implies LP = MP$.
Since $AL = CM \implies AL + LP = CM + MP \implies \mathbf{AP = CP}$.
Subtracting: $AB - AP = CD - CP \implies \mathbf{BP = DP}$. ■

Fig 5.5A: Intersecting Chords

Topic 6: Angles Subtended by an Arc & Cyclic Theorems

Theorem 9: Central Angle is Double Circumference Angle

Statement: The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

$$\mathbf{\angle AOB = 2 \angle ACB}$$

Fig 5.11: Proof Diagram for Theorem 9 — Central Angle $\angle AOB = 2 \angle ACB$

Proof: In $\Delta AOC$, $OA = OC = r \implies \angle AOE = 2 \angle OCA$. Similarly $\angle BOE = 2 \angle OCB$. Adding: $\mathbf{\angle AOB = 2 \angle ACB}$. ■

Topic 6: School Exam Practice Problem Kit

Problem 6.1 (Reflex Angle Arc Calculation) 3 MARKS

In a circle with centre $O$, major arc $AB$ subtends a reflex central angle $\angle AOB = 250^\circ$. Find $\angle ACB$ where $C$ is a point on the minor arc.

Step-by-Step Solution:

$$\angle ACB = \frac{\text{Reflex Angle}}{2} = \frac{250^\circ}{2} = \mathbf{125^\circ}$$

Fig 5.6A: Reflex Central Angle

Topic 7: Concyclicity of Points & Cyclic Quadrilaterals

Concyclicity Rules

Concyclic Points: Four or more points lying on the same circle.
Theorem 10: If a line segment joining two points subtends equal angles at two other points on the same side, the 4 points are concyclic.
Cyclic Quadrilateral: A quadrilateral whose all 4 vertices lie on a circle.

Theorem 11: Sum of Opposite Angles of Cyclic Quad is 180°

Statement: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.

$$\mathbf{\angle A + \angle C = 180^\circ} \qquad \mathbf{\angle B + \angle D = 180^\circ}$$

Fig 5.14: Proof Diagram for Theorem 11 — Cyclic Quadrilateral $ABCD$

Proof: By Theorem 9, $\angle BAD = \frac{1}{2}(\text{Reflex } \angle BOD)$ and $\angle BCD = \frac{1}{2}(\text{Angle } \angle BOD)$. Adding both equations gives $\frac{1}{2}(360^\circ) = \mathbf{180^\circ}$. ■

Topic 7: School Exam Practice Problem Kit

Problem 7.1 (Exterior Angle Property) 2 MARKS

If side $BC$ of a cyclic quadrilateral $ABCD$ is produced to point $E$, prove that exterior angle $\angle DCE$ equals interior opposite angle $\angle BAD$.

Step-by-Step Solution:

$\angle BCD + \angle DCE = 180^\circ$ (linear pair) and $\angle BAD + \angle BCD = 180^\circ$.
Equating both: $\mathbf{\angle DCE = \angle BAD}$. ■

Fig 5.7A: Exterior Angle $\angle DCE$

Topic 8: Comprehensive Exam HOTS & Case Study Kit

Case Study 1: Parallel Chords Geometry CASE STUDY

Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ are on the same side of the centre of a circle. The distance between the chords is $7\text{ cm}$. Find the radius $r$ of the circle.

Fig 5.15: Case Study 1 — Parallel Chords ($r = 13\text{ cm}$)

Solution:

Let distance of $24\text{ cm}$ chord be $x \implies r^2 = x^2 + 12^2 = x^2 + 144$.

Distance of $10\text{ cm}$ chord $= x + 7 \implies r^2 = (x+7)^2 + 5^2 = x^2 + 14x + 74$.

Equating: $x^2 + 144 = x^2 + 14x + 74 \implies 14x = 70 \implies x = 5\text{ cm} \implies r = \sqrt{25 + 144} = \mathbf{13\text{ cm}}$.

9. 100/100 School Examination Master Checklist

Master Exam Checklist

Unique Circle: Exactly 1 unique circle passes through 3 non-collinear points. Zero through 3 collinear points.
Equal Chords: Equal chords subtend equal central angles and lie at equal distances from centre.
Perpendicular Bisector: Perpendicular from centre bisects chord ($r^2 = d^2 + (L/2)^2$).
Angle at Centre: Central angle is double circumference angle ($\angle AOB = 2 \angle ACB$).
Semicircle Angle: Angle in a semicircle is a right angle ($90^\circ$).
Same Segment Angles: Angles subtended in the same segment of a circle are equal.
Cyclic Quad: Opposite angles sum to $180^\circ$. Exterior angle $=$ Interior opposite angle. Cyclic parallelogram is a rectangle.