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ICSE • Class 9 • Mathematics • Ch 16
Estimated Time: 45 Mins
Study Progress: In Progress

Circle

In ICSE Class 9 Mathematics, "Circle" introduces the formal geometric study of loci of points equidistant from a fixed center in a plane, establishing the core chord and arc theorems of Euclidean geometry. A circle $C(O, r)$ is formally defined as the set of all points in a plane whose distance from a fixed point $O$ (center) is equal to a constant non-negative real radius $r$. The curriculum thoroughly develops five landmark chord theorems: (1) The perpendicular drawn from the center of a circle to a chord bisects the chord; (2) The converse theorem: The line joining the center of a circle to the mid-point of a chord is perpendicular to the chord; (3) There is one and only one circle passing through three non-collinear points (circumcircle construction via perpendicular bisectors); (4) Equal chords of a circle are equidistant from the center; and (5) Chords equidistant from the center of a circle are equal in length. The chapter further explores arc congruency and degree measure: equal arcs subtend equal angles at the center and equal chords, and congruent arcs correspond to congruent central angles. Rigorous algebraic calculations using the Pythagorean relation $r^2 = d^2 + \left(\frac{L}{2}\right)^2$ (where $r$ is the radius, $d$ is perpendicular distance from center, and $L$ is chord length) are mastered for intersecting chords and parallel chords on the same or opposite sides of the center.

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The Archeologist's Broken Plate: How Three Random Shards Can Reconstruct an Ancient Roman Circle

Imagine you are an archaeologist digging in Pompeii and you unearth a broken fragment of an ancient Roman ceremonial circular bronze shield. The center of the shield was lost two thousand years ago in volcanic ash. How can you determine the exact original radius of the shield so museum curators can reconstruct it? Do you need the center point? No! Euclid's third circle theorem proves that any three non-collinear points uniquely determine one and only one circle in the universe! By selecting any three tiny rim points $A, B, C$ on the broken rim, drawing chords $AB$ and $BC$, and finding where their perpendicular bisectors intersect, you pinpoint the exact center $O$ of the circle! Why does the perpendicular from the center always slice every chord into two identical halves? How do parallel chords across the center reveal the secrets of circular symmetry? Let us master the geometry of circles!

Why This Chapter Matters

Circle theorems govern civil engineering (circular bridge arches, roadway roundabouts), planetary orbits, satellite radar cones, gears and mechanical transmission, and optics.

Before You Begin (Prerequisites)

  • Congruence criteria (SSS, SAS, RHS) from Chapter 8.
  • Pythagoras Theorem ($a^2 + b^2 = c^2$) from Chapter 12.

What You Will Learn (Core Objectives)

  • State and prove: Perpendicular from center to a chord bisects the chord.
  • State and prove the converse: Line joining center to midpoint of chord is perpendicular.
  • Prove that there is one and only one circle passing through three non-collinear points.
  • Prove: Equal chords are equidistant from the center, and converse.
  • Calculate chord lengths, radii, and center distances using Pythagorean relations.

Chapter Roadmap & Progression

1 1. Perpendicular from Center to a C...
2 2. The Pythagorean Chord Metric Rel...
3 3. Equal Chords and Distance from C...
4 4. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Perpendicular from Center to a Chord & Converse

Fundamental Chord Theorems
Theorem 1:

Statement: The perpendicular from the center of a circle to a chord bisects the chord.

Proof:

Let $AB$ be a chord of a circle with center $O$. Draw $OM \perp AB$. We must prove $AM = MB$.

Join $OA$ and $OB$. In right-angled triangles $\triangle OMA$ and $\triangle OMB$:

  1. $OA = OB$ (Radii of the same circle, Hypotenuse).
  2. $OM = OM$ (Common side).
  3. $\angle OMA = \angle OMB = 90^\circ$ (Right angle).

By RHS Congruence Criterion: $\triangle OMA \cong \triangle OMB$.

By CPCTC: $\mathbf{AM = MB}$. $\blacksquare$

Theorem 2 (Converse):

Statement: The line joining the center of a circle to the mid-point of a chord is perpendicular to the chord.

Proof: In $\triangle OMA$ and $\triangle OMB$, $OA = OB$ (radii), $AM = MB$ (given), $OM = OM$ (common). By SSS: $\triangle OMA \cong \triangle OMB \implies \angle OMA = \angle OMB$. Since they form a linear pair: $\angle OMA = \frac{180^\circ}{2} = 90^\circ \implies \mathbf{OM \perp AB}$. $\blacksquare$

2. The Pythagorean Chord Metric Relation

Quantitative Chord Calculation

In right-angled triangle $\triangle OMA$:

$$\mathbf{r^2 = d^2 + \left(\frac{L}{2}\right)^2}$$
  • $r = OA$ is the radius of the circle.
  • $d = OM$ is the perpendicular distance from center $O$ to chord $AB$.
  • $L = AB$ is the length of the chord, so $AM = \frac{L}{2}$.
  • Chord length: $\mathbf{L = 2\sqrt{r^2 - d^2}}$
  • Perpendicular distance: $\mathbf{d = \sqrt{r^2 - \left(\frac{L}{2}\right)^2}}$

3. Equal Chords and Distance from Center

Distance Theorems
Theorem 3:

Statement: Equal chords of a circle (or of congruent circles) are equidistant from the center.

$$\mathbf{AB = CD \implies OM = ON}$$

Proof Sketch: In right triangles $\triangle OMA$ and $\triangle ONC$: $OA = OC = r$, and $AM = \frac{1}{2}AB = \frac{1}{2}CD = CN$. By RHS: $\triangle OMA \cong \triangle ONC \implies OM = ON$.

Theorem 4 (Converse):

Statement: Chords equidistant from the center of a circle are equal in length.

$$\mathbf{OM = ON \implies AB = CD}$$

4. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ are drawn on opposite sides of the center of a circle of radius $13\text{ cm}$. Find the distance between the two chords.

Solution:

Let the two parallel chords be $AB = 10\text{ cm}$ and $CD = 24\text{ cm}$. Let $O$ be the center, and radius $r = 13\text{ cm}$.

Draw $OM \perp AB$ and $ON \perp CD$. Since $AB \parallel CD$ and they lie on opposite sides of $O$, the points $M, O, N$ are collinear.

The total distance between chords is $MN = OM + ON$.

1. In right-angled $\triangle OMA$:

$$AM = \frac{AB}{2} = \frac{10}{2} = 5\text{ cm}$$ $$OM = \sqrt{r^2 - AM^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$$

2. In right-angled $\triangle ONC$:

$$CN = \frac{CD}{2} = \frac{24}{2} = 12\text{ cm}$$ $$ON = \sqrt{r^2 - CN^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}$$

3. Distance between the parallel chords:

$$MN = OM + ON = 12 + 5 = \mathbf{17\text{ cm}}$$

Note: If the chords were on the same side of the center, the distance would be $OM - ON = 12 - 5 = 7\text{ cm}$.

Key Formulas, Identities & Theorems

Chord Metric Formula
$$r^2 = d^2 + \left(\frac{L}{2}\right)^2$$
Relates radius r, perpendicular distance d, and chord length L.
Chord Length from Distance
$$L = 2\sqrt{r^2 - d^2}$$
Calculates chord length.
Equal Chords Equidistance
$$AB = CD \iff OM = ON$$
Equal chords are equidistant from the center.

Mathematics: Chord Bisector Theorem & Parallel Chords Metric

Circle Chord Theorems: Perpendicular Bisection & Parallel Chords OM ⊥ AB ⇒ AM = MB O A B M r r ΔOMA ≅ ΔOMB (RHS) ⇒ AM = MB Metric: r² = d² + (L/2)² Two Parallel Chords on Opposite Sides O AB CD d₁=12 d₂=5 Total Distance = d₁ + d₂ = 12 + 5 = 17 cm

Chapter Summary & 10 Key Takeaways

Takeaway 1
A circle is the planar locus of points equidistant from a fixed center.
Takeaway 2
The perpendicular from the center of a circle to a chord bisects the chord.
Takeaway 3
The line joining the center to the midpoint of a chord is perpendicular to the chord.
Takeaway 4
The fundamental metric relation is r^2 = d^2 + (L/2)^2.
Takeaway 5
There is one and only one circle passing through three non-collinear points.
Takeaway 6
Equal chords of a circle are equidistant from the center.
Takeaway 7
Chords equidistant from the center of a circle are equal in length.
Takeaway 8
For two parallel chords on opposite sides of the center, distance is d1 + d2; on the same side, distance is |d1 - d2|.
Takeaway 9
An arc of a circle is a continuous portion of the circumference.
Takeaway 10
Equal arcs subtend equal angles at the center and correspond to equal chords.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A chord of length $16\text{ cm}$ is at a distance of $6\text{ cm}$ from the center of a circle. Find the radius of the circle.
Reveal Answer & Explanation
Answer: • Given: Chord length $L = 16\text{ cm}$, Perpendicular distance $d = 6\text{ cm}$.
• The perpendicular from the center bisects the chord, so half the chord length is:
$$AM = \frac{L}{2} = \frac{16}{2} = 8\text{ cm}$$
• In right-angled triangle $\triangle OMA$, by the Pythagoras Theorem:
$$r^2 = d^2 + AM^2$$
$$r^2 = 6^2 + 8^2 = 36 + 64 = 100$$
$$r = \sqrt{100} = \mathbf{10\text{ cm}}$$
r = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cm.
2
Find the length of a chord which is at a distance of $5\text{ cm}$ from the center of a circle of radius $13\text{ cm}$.
Reveal Answer & Explanation
Answer: • Given: Radius $r = 13\text{ cm}$, distance $d = 5\text{ cm}$.
• Using the metric relation $AM = \sqrt{r^2 - d^2}$:
$$AM = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$$
• The full chord length is twice $AM$:
$$L = 2 \times AM = 2 \times 12 = \mathbf{24\text{ cm}}$$
Half-chord = √(13^2 - 5^2) = 12 cm. Full chord = 2 * 12 = 24 cm.
3
Two concentric circles have radii $5\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.
Reveal Answer & Explanation
Answer: • Let the common center be $O$.
• A chord $AB$ of the larger circle ($R = 5\text{ cm}$) touches the smaller circle ($r = 3\text{ cm}$) at point $M$.
• The radius to the point of contact is perpendicular to the tangent chord: $OM \perp AB$, so $OM = 3\text{ cm}$.
• In right-angled $\triangle OMA$ with hypotenuse $OA = R = 5\text{ cm}$ and leg $OM = 3\text{ cm}$:
$$AM = \sqrt{OA^2 - OM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$
• Since $OM \perp AB$, $M$ bisects $AB$:
$$AB = 2 \times AM = 2 \times 4 = \mathbf{8\text{ cm}}$$
OM = 3 cm, OA = 5 cm. AM = √(25 - 9) = 4 cm. Full chord = 2 * 4 = 8 cm.
4
Prove that there is one and only one circle passing through three given non-collinear points.
Reveal Answer & Explanation
Answer:

• Let $A, B, C$ be three non-collinear points.
• Join $AB$ and $BC$.
• Draw the perpendicular bisector $l_1$ of segment $AB$ and the perpendicular bisector $l_2$ of segment $BC$.
• Since $A, B, C$ are non-collinear, the lines $AB$ and $BC$ are not parallel; hence their perpendicular bisectors $l_1$ and $l_2$ are not parallel and must intersect at exactly one unique point $O$.
• Since $O$ lies on $l_1$, $OA = OB$. Since $O$ lies on $l_2$, $OB = OC$.
• Therefore, $OA = OB = OC = r$.
• A circle with center $O$ and radius $r$ will pass through $A, B$, and $C$.
• Since $l_1$ and $l_2$ intersect at only one point $O$, there is one and only one such circle. $\blacksquare$


Perpendicular bisectors of AB and BC intersect at a single unique center point O where OA = OB = OC.
5
Two parallel chords of a circle of radius $5\text{ cm}$ are of lengths $8\text{ cm}$ and $6\text{ cm}$. If the chords are on the SAME side of the center, find the distance between them.
Reveal Answer & Explanation
Answer:

• Given $r = 5\text{ cm}$, Chords $AB = 8\text{ cm}$ and $CD = 6\text{ cm}$.
• 1. Distance to chord $AB$ ($L_1 = 8\text{ cm}$, half $= 4\text{ cm}$):

$$d_1 = \sqrt{r^2 - 4^2} = \sqrt{5^2 - 16} = \sqrt{9} = 3\text{ cm}$$


• 2. Distance to chord $CD$ ($L_2 = 6\text{ cm}$, half $= 3\text{ cm}$):

$$d_2 = \sqrt{r^2 - 3^2} = \sqrt{5^2 - 9} = \sqrt{16} = 4\text{ cm}$$


• Since both chords lie on the same side of the center:

$$\text{Distance between chords} = d_2 - d_1 = 4 - 3 = \mathbf{1\text{ cm}}$$


d1 = √(25 - 16) = 3 cm, d2 = √(25 - 9) = 4 cm. On the same side: distance is 4 - 3 = 1 cm.
6
Prove that equal chords of a circle subtend equal angles at the center.
Reveal Answer & Explanation
Answer:

• Let $AB$ and $CD$ be two equal chords of a circle with center $O$ ($AB = CD$).
• In $\triangle AOB$ and $\triangle COD$:
1. $OA = OC$ (Radii of the same circle).
2. $OB = OD$ (Radii of the same circle).
3. $AB = CD$ (Given).
• By SSS Congruence Criterion: $\triangle AOB \cong \triangle COD$.
• By CPCTC: $\mathbf{\angle AOB = \angle COD}$. $\blacksquare$


Use SSS on triangles AOB and COD with all radii equal and AB = CD.
7
If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.
Reveal Answer & Explanation
Answer:

• Let $AB$ and $CD$ be two chords intersecting at point $E$ inside the circle. Let $PQ$ be the diameter through $E$ such that $\angle AEP = \angle DEP$.
• Draw perpendiculars from center $O$ to both chords: $OL \perp AB$ and $OM \perp CD$.
• In $\triangle OLE$ and $\triangle OME$:
1. $\angle OLE = \angle OME = 90^\circ$ (By construction).
2. $\angle OEL = \angle OEM$ (Given, equal angles with diameter).
3. $OE = OE$ (Common hypotenuse).
• By AAS Congruence Criterion: $\triangle OLE \cong \triangle OME$.
• By CPCTC: $OL = OM$.
• Since chords equidistant from the center are equal in length (Theorem 4):

$$\mathbf{AB = CD} \quad \blacksquare$$


Draw perpendiculars OL and OM from center. Use AAS to prove OL = OM, which implies chords are equal.
8
Why can a circle NOT pass through three collinear points?
Reveal Answer & Explanation
Answer:

• If points $A, B, C$ are collinear (lie on a single straight line), then segments $AB$ and $BC$ lie on the same line.
• The perpendicular bisectors of $AB$ and $BC$ are both perpendicular to the same straight line.
• Two distinct lines perpendicular to the same straight line are strictly parallel to each other and never intersect.
• Since the perpendicular bisectors have no point of intersection, there is no equidistant center point $O$. Hence, no circle can pass through three collinear points.


Perpendicular bisectors of segments on a single line are parallel and never intersect.
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