Circles

This chapter provides a rigorous treatment of fundamental circle properties, encompassing angles, chords, tangents, and secants. Proficiency in these concepts is essential for CMI examinations, where problems frequently integrate topics such as power of a point and circle-based similarity for advanced geometric analysis.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Angle in a circle | | 2 | Chords | | 3 | Tangents | | 4 | Secants | | 5 | Power of a point | | 6 | Circle-based similarity problems |

---

We begin with Angle in a circle.

Part 1: Angle in a circle

Angle in a Circle

Overview

Angles in a circle form one of the most important theorem clusters in Euclidean geometry. These results connect chords, arcs, central angles, inscribed angles, semicircles, and cyclic quadrilaterals. In exam problems, the essential skill is to identify which angles stand on the same chord or same arc. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Use the inscribed-angle theorem correctly.

Relate central angles and angles at the circumference.

Recognize when angles in the same segment are equal.

Use the angle-in-a-semicircle result effectively.

Solve cyclic quadrilateral and chord-based angle problems.

---

Core Theorems

📐 Angle at the Centre and Circumference

If $A$ and $B$ are points on a circle with centre $O$ , and $P$ is another point on the same circle on the same side of chord $AB$ , then

$\qquad \angle AOB = 2\angle APB$

So the angle subtended by a chord at the centre is twice the angle subtended by the same chord at the circumference.

📐 Angles in the Same Segment

Angles standing on the same chord and lying in the same segment are equal.

So if $A,B,C,D$ lie on a circle on the same side of chord $AB$ , then

$\qquad \angle ACB = \angle ADB$

📐 Angle in a Semicircle

If $AB$ is a diameter of a circle and $P$ is any point on the circle, then

$\qquad \angle APB = 90^\circ$

This is one of the most frequently used theorems in school and olympiad geometry. ---

Cyclic Quadrilateral Angle Facts

📐 Opposite Angles in a Cyclic Quadrilateral

If $ABCD$ is cyclic, then

$\qquad \angle A + \angle C = 180^\circ$

$\qquad \angle B + \angle D = 180^\circ$

❗ Converse

If in a quadrilateral a pair of opposite angles is supplementary, then the quadrilateral is cyclic.

---

Why Same Chord Matters

📖 Standing on the Same Chord

When we say an angle stands on chord $AB$ , we mean its arms pass through $A$ and $B$ .

So:

$\angle AOB$ stands on chord $AB$

$\angle APB$ stands on chord $AB$

$\angle AQB$ stands on chord $AB$

This is why those angles are connected.

---

Minimal Worked Examples

Example 1 If a chord subtends an angle of

64^\circ

at the circumference, then the angle subtended by the same chord at the centre is

\qquad 2\times 64^\circ = 128^\circ

--- Example 2 If

AB

is a diameter and

P

lies on the circle, then

\qquad \angle APB = 90^\circ

This often creates a hidden right triangle. ---

Common Standard Deductions

📐 Quick Deductions

same chord $\Rightarrow$ same inscribed angle

centre angle $=$ twice circumference angle on same chord

diameter $\Rightarrow$ right angle at the circumference

cyclic quadrilateral $\Rightarrow$ opposite angles supplementary

---

Hidden Structure in Problems

💡 What to Look For

When a problem involves many angles on a circle, immediately check:

do two angles stand on the same chord?

is there a diameter?

is a quadrilateral cyclic?

can a centre angle be converted into an inscribed angle?

---

Common Mistakes

⚠️ Avoid These Errors

❌ Using same-segment equality for angles standing on different chords

❌ Forgetting that the centre angle is twice, not equal to, the inscribed angle

❌ Applying angle-in-a-semicircle when the subtended chord is not a diameter

❌ Forgetting the converse of cyclic quadrilateral when opposite angles are supplementary

---

CMI Strategy

💡 How to Attack Circle-Angle Problems

Mark the chord on which each angle stands.

Convert centre angles into circumference angles when useful.

Search for a diameter immediately.

If four points lie on a circle, check opposite angles.

Reduce the problem to one known theorem at a time.

---

Practice Questions

:::question type="MCQ" question="A chord of a circle subtends an angle of

35^\circ

at a point on the circumference. The angle subtended by the same chord at the centre is" options=["

35^\circ

","

70^\circ

","

105^\circ

","

140^\circ

"] answer="B" hint="Use the centre-angle theorem." solution="The angle subtended at the centre is twice the angle subtended at the circumference on the same chord. Therefore it is

\qquad 2\times 35^\circ = 70^\circ

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="If

AB

is a diameter of a circle and

P

is a point on the circle, find

\angle APB

in degrees." answer="90" hint="Use the angle-in-a-semicircle theorem." solution="Since

AB

is a diameter, the angle subtended by

AB

at the circumference is a right angle. Therefore

\qquad \angle APB = 90^\circ

. Hence the answer is

\boxed{90}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Angles in the same segment are equal","The angle subtended by a chord at the centre is twice the angle subtended at the circumference on the same chord","Opposite angles of a cyclic quadrilateral are supplementary","Every quadrilateral with one right angle is cyclic"] answer="A,B,C" hint="Recall the standard circle theorems." solution="1. True. This is the same-segment theorem.

True. This is the centre-angle theorem.

True. This is a standard property of cyclic quadrilaterals.

False. One right angle alone does not make a quadrilateral cyclic.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Prove that the angle subtended by a diameter at the circumference is a right angle." answer="True" hint="Use the theorem that the centre angle is twice the circumference angle." solution="Let

AB

be a diameter of a circle with centre

O

, and let

P

be any other point on the circle. Since

AB

is a diameter, the angle at the centre standing on chord

AB

\qquad \angle AOB = 180^\circ

. By the theorem that the angle at the centre is twice the angle at the circumference on the same chord,

\qquad \angle AOB = 2\angle APB

. So

\qquad 180^\circ = 2\angle APB

, which gives

\qquad \angle APB = 90^\circ

. Therefore the angle subtended by a diameter at the circumference is a right angle. Hence the answer is

\boxed{\text{True}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Centre angle is twice the inscribed angle on the same chord.

Angles in the same segment are equal.

A diameter subtends a right angle at the circumference.

Opposite angles of a cyclic quadrilateral are supplementary.

In circle-angle problems, identifying the correct chord is usually the main step.

---

💡 Next Up

Proceeding to Chords.

---

Part 2: Chords

Chords

Overview

A chord is a line segment joining two points on a circle. Chord questions in Euclidean geometry look simple, but they often hide strong relations between the center, the distance from the center, equal chords, and angles subtended by the same chord. In exam problems, the main skill is to combine these standard theorems quickly and cleanly. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

use the fundamental properties of chords in a circle,

relate chords to distance from the center,

prove equality of chords and angles,

identify midpoint and perpendicular relations,

solve geometry problems involving chord lengths and central structure.

---

Core Idea

📖 Chord

A chord of a circle is a line segment whose endpoints lie on the circle.

A diameter is a special chord passing through the center.

---

Fundamental Chord Theorems

📐 Perpendicular from Center to a Chord

If a perpendicular from the center of a circle meets a chord, then it bisects the chord.

So if $O$ is the center and $OM\perp AB$ , then

$\qquad AM=MB$

📐 Converse

If a line from the center passes through the midpoint of a chord, then it is perpendicular to the chord.

So if $M$ is the midpoint of chord $AB$ , then

$\qquad OM\perp AB$

These two results are used constantly. ---

Equal Chords and Equal Distances

📐 Equal Chords

In the same circle, or in equal circles:

equal chords are equidistant from the center,

chords equidistant from the center are equal.

So if chords

AB

and

CD

satisfy

\qquad d(O,AB)=d(O,CD)

then

\qquad AB=CD

and conversely.

📐 Longer Chord is Closer to the Center

In the same circle:

if one chord is longer than another, then it is nearer to the center,

if one chord is nearer to the center, then it is longer.

This is an important ordering theorem. ---

Chords and Angles

📐 Equal Chords Subtend Equal Angles at the Center

If chords $AB$ and $CD$ are equal in the same circle, then the central angles are equal:

$\qquad AB=CD \implies \angle AOB=\angle COD$

📐 Equal Chords Subtend Equal Angles at the Circumference

Equal chords also subtend equal angles at any point on the same segment of the circle.

This lets you convert chord equality into angle equality.

❗ Same Chord, Same Segment

Angles subtended by the same chord in the same segment are equal.

If points $P$ and $Q$ lie on the same segment of chord $AB$ , then

$\qquad \angle APB=\angle AQB$

This is one of the main angle tools in circle geometry. ---

Chord Bisector Structure

📐 Line Joining Center to Midpoint

If $M$ is the midpoint of chord $AB$ and $O$ is the center, then

$\qquad OA=OB,\quad AM=MB,\quad OM\perp AB$

So triangles $\triangle OAM$ and $\triangle OBM$ are congruent.

This gives a standard right-triangle setup for many chord-length problems. ---

Chord Length Formula

📐 Chord Length from Radius and Distance

If a circle has radius $r$ , and a chord is at perpendicular distance $d$ from the center, then half the chord length is

$\qquad \sqrt{r^2-d^2}$

So the full chord length is

$\qquad 2\sqrt{r^2-d^2}$

This follows from Pythagoras in the right triangle formed by the radius, half-chord, and perpendicular distance. ---

Minimal Worked Examples

Example 1 A circle has radius

5

, and a chord is at distance

3

from the center. Find its length. Half the chord is

\qquad \sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4

Therefore the chord length is

\qquad 2\times 4=8

So the required length is

\boxed{8}

. --- Example 2 Two chords of the same circle are at equal distances from the center. What can you conclude? Since chords equidistant from the center are equal, the two chords must have equal length. ---

Common Patterns

📐 What Exam Questions Usually Test

midpoint of a chord,

perpendicular from center to chord,

comparing two chords using distances from center,

finding chord length using radius,

converting chord equality into angle equality.

---

Common Mistakes

⚠️ Avoid These Errors

❌ assuming every line from the center to a chord is perpendicular,

✅ this is true only when it passes through the midpoint

❌ confusing chord with arc,

✅ a chord is a segment, not the curved part

❌ forgetting that theorems are for the same circle or equal circles,

✅ check the geometric setting carefully

❌ using angle theorems without checking same segment or same chord,

✅ state the chord or arc explicitly

---

CMI Strategy

💡 How to Solve Chord Problems

Mark the center first.

Look for midpoint and perpendicular information.

Use congruent right triangles whenever possible.

Translate equal chords into equal distances or equal angles.

In computation problems, reduce everything to Pythagoras.

---

Practice Questions

:::question type="MCQ" question="In a circle with center

O

, if

OM

is perpendicular to chord

AB

, then which of the following must be true?" options=["

OA=OM

","

AM=MB

","

AB

is a diameter","

OM=AB

"] answer="B" hint="Use the perpendicular from center theorem." solution="A perpendicular from the center of a circle to a chord bisects the chord. Therefore if

OM\perp AB

, then

\boxed{AM=MB}

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="A circle has radius

10

. A chord is at distance

6

from the center. Find the length of the chord." answer="16" hint="Use the right triangle with half-chord." solution="If the radius is

10

and the perpendicular distance from the center to the chord is

6

, then half the chord is

\qquad \sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8

Therefore the full chord length is

\qquad 2\times 8=16

Hence the answer is

\boxed{16}

." ::: :::question type="MSQ" question="Which of the following statements are true in the same circle?" options=["Equal chords are equidistant from the center","A longer chord is nearer to the center","Every radius is perpendicular to every chord","If a line from the center bisects a chord, then it is perpendicular to the chord"] answer="A,B,D" hint="Recall the basic chord theorems." solution="1. True. Equal chords are equidistant from the center.

True. Among chords in the same circle, the longer chord lies nearer to the center.

False. A radius is perpendicular to a chord only in special cases.

True. This is the converse midpoint theorem for chords.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Prove that chords equidistant from the center of a circle are equal." answer="Using congruent right triangles formed by perpendiculars from the center, the half-chords are equal, hence the full chords are equal" hint="Drop perpendiculars from the center to both chords." solution="Let

AB

and

CD

be chords of a circle with center

O

, and suppose their perpendicular distances from the center are equal. Let

OM\perp AB

and

ON\perp CD

, with

OM=ON

. By the theorem on perpendicular from the center to a chord, we have

\qquad AM=MB

and

\qquad CN=ND

Now

OA=OC

since both are radii, and

OM=ON

by assumption. So in right triangles

\triangle OAM

and

\triangle OCN

, we have:

$OA=OC$

$OM=ON$

both are right triangles

Hence the triangles are congruent, so

\qquad AM=CN

Therefore

\qquad AB=2AM=2CN=CD

Thus chords equidistant from the center are equal." ::: ---

Summary

❗ Key Takeaways for CMI

A perpendicular from the center to a chord bisects the chord.

The converse is also true.

Equal chords correspond to equal distances from the center.

Longer chords lie nearer to the center.

Chord problems often reduce to right triangles and symmetry.

---

💡 Next Up

Proceeding to Tangents.

---

Part 3: Tangents

Tangents

Overview

A tangent to a circle is a line that touches the circle at exactly one point. Tangent problems in Euclidean geometry are important because they connect angle facts, power of a point, equal lengths, and perpendicularity. In exam questions, the main skill is to identify the tangent point and then use the correct standard property immediately. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

State and use the basic properties of tangents to a circle.

Relate tangents to radii and chords.

Use equal tangent lengths from an external point.

Solve angle problems involving tangents and circles.

Recognise tangent-based configurations quickly in exam questions.

---

Core Definition

📖 Tangent to a Circle

A tangent to a circle is a line that meets the circle at exactly one point.

That point is called the point of contact or point of tangency.

---

Most Important Tangent Property

📐 Radius-Tangent Perpendicularity

If a line is tangent to a circle at a point $P$ , and $O$ is the centre of the circle, then

$\qquad OP \perp \text{tangent at } P$

❗ Converse Form

If a line through a point $P$ on a circle is perpendicular to the radius $OP$ , then that line is tangent to the circle at $P$ .

This is one of the most used facts in geometry problems. ---

Equal Tangents from an External Point

📐 Equal Tangent Lengths

If from an external point $A$ , tangents $AB$ and $AC$ are drawn to a circle touching it at $B$ and $C$ , then

$\qquad AB = AC$

❗ Useful Consequences

In the same configuration:

$\triangle AOB \cong \triangle AOC$

$AO$ bisects $\angle BAC$

$AO$ is the perpendicular bisector of chord $BC$

These facts follow from radius equality and right-triangle congruence. ---

Angle Between Tangent and Radius

📐 At the Point of Contact

The angle between a tangent and the radius through the point of contact is

$\qquad 90^\circ$

This is usually the first angle fact to mark in a diagram. ---

Tangent-Chord Angle Theorem

📐 Angle in Alternate Segment

The angle between a tangent and a chord through the point of contact is equal to the angle subtended by that chord in the opposite arc.

If $PT$ is tangent at $P$ and $PQ$ is a chord, then

$\qquad \angle(\text{tangent } PT,\ \text{chord } PQ) = \angle PRQ$

where $R$ is a point on the opposite arc.

❗ Exam Meaning

Whenever you see:

one tangent

one chord through the point of contact

one angle on the circle

think of the alternate-segment theorem immediately.

---

Tangent and Secant Length Relation

📐 Power of a Point: Tangent-Secant Form

If from an external point $A$ , a tangent $AT$ and a secant through points $B,C$ are drawn to a circle, then

$\qquad AT^2 = AB \cdot AC$

This is very useful in length problems. ---

Tangents to Two Circles

💡 Common Tangent Intuition

For two circles, a common tangent touches both circles. In standard school-level problems:

radii to tangent points are perpendicular to the common tangent

distances involving centres often create right triangles

external and internal common tangents may behave differently

---

Standard Configurations

💡 Patterns to Recognise

Radius drawn to point of contact

Two tangents from one external point

Tangent and chord at same point

Tangent and secant from same external point

Tangent used inside a cyclic quadrilateral problem

---

Minimal Worked Examples

Example 1 From an external point

A

, tangents

AB

and

AC

are drawn to a circle. If

AB=7

, then

\qquad AC=7

because tangent lengths from the same external point are equal. --- Example 2 A radius to the point of contact has length

5

. The tangent through that point is perpendicular to the radius. So the angle between them is

\qquad 90^\circ

---

Common Mistakes

⚠️ Avoid These Errors

❌ Thinking a tangent can cut the circle at two points

✅ A tangent touches the circle at exactly one point

❌ Forgetting that the radius to the point of contact is perpendicular to the tangent

✅ Always mark the right angle first

❌ Using equal tangent lengths from two different external points

✅ Equality holds only for tangents from the same external point

❌ Mixing chord-angle facts with tangent-angle facts

✅ Identify the exact theorem before using it

---

CMI Strategy

💡 How to Attack Tangent Problems

Mark the point of contact clearly.

Draw the radius to that point immediately.

Use the $90^\circ$ fact first.

If two tangents come from one point, write equal lengths at once.

If a tangent and a secant appear from the same point, test whether power of a point applies.

If a tangent and a chord meet, think alternate-segment theorem.

---

Practice Questions

:::question type="MCQ" question="If a line is tangent to a circle at

P

and

O

is the centre, then

\angle OP\text{(tangent)}

equals" options=["

0^\circ

","

45^\circ

","

90^\circ

","

180^\circ

"] answer="C" hint="Use the radius-tangent property." solution="The radius drawn to the point of contact is perpendicular to the tangent. Therefore the angle is

\boxed{90^\circ}

, so the correct option is

\boxed{C}

." ::: :::question type="NAT" question="From an external point

A

, tangents

AB

and

AC

are drawn to a circle. If

AB=12

, find

AC

." answer="12" hint="Tangents from the same external point have equal lengths." solution="Tangents drawn from the same external point to a circle are equal in length. Hence

\qquad AC = AB = 12

. Therefore the answer is

\boxed{12}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["A radius to the point of contact is perpendicular to the tangent","Two tangents from the same external point are equal in length","A tangent always passes through the centre of the circle","The angle between a tangent and a chord can be related to an angle in the opposite arc"] answer="A,B,D" hint="Recall the standard tangent theorems." solution="1. True. This is the basic tangent-radius theorem.

True. Tangents from the same external point are equal.

False. A tangent does not pass through the centre in general.

True. This is the tangent-chord angle theorem.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Explain why tangents drawn from the same external point to a circle are equal in length." answer="Because the triangles formed with the centre and the tangent points are congruent right triangles." hint="Join the centre to the tangent points." solution="Let from an external point

A

, tangents

AB

and

AC

be drawn to a circle with centre

O

, touching the circle at

B

and

C

. Join

OB

OC

, and

OA

. Since a radius to the point of contact is perpendicular to the tangent,

\angle OBA = \angle OCA = 90^\circ

. Also,

OB = OC

because they are radii, and

OA

is common. Therefore the right triangles

\triangle OBA

and

\triangle OCA

are congruent by RHS. Hence corresponding sides are equal, so

\qquad AB = AC

." ::: ---

Summary

❗ Key Takeaways for CMI

A radius to the point of contact is perpendicular to the tangent.

Tangents from the same external point are equal.

Tangent-chord angle equals the angle in the opposite arc.

Tangent-secant problems often use power of a point.

In most tangent problems, the right angle at the point of contact is the first crucial step.

---

💡 Next Up

Proceeding to Secants.

---

Part 4: Secants

Secants

Overview

A secant is a line that cuts a circle at two points. This simple object appears in many geometry questions involving intercepted arcs, external angles, and segment products. In CMI-style problems, secants are often used together with arc reasoning and power-of-a-point ideas. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

identify secants and distinguish them from chords and tangents,

use angle relations involving secants,

apply secant-secant reasoning from an external point,

convert circle diagrams into arc and angle equations,

solve secant-based length and angle problems efficiently.

---

Core Definitions

📖 Secant and Related Objects

A secant is a line that meets a circle at two distinct points.

A chord is the segment joining two points on the circle.

A tangent is a line that touches the circle at exactly one point.

❗ Useful Distinction

A chord lies inside the circle, while a secant is the full line through the circle.

---

Secants and Exterior Angles

📐 Angle Formed by Two Secants Outside a Circle

If two secants from an external point form an angle, then the measure of that angle equals half the difference of the intercepted arcs.

$\qquad \angle = \dfrac{1}{2}(\text{larger intercepted arc} - \text{smaller intercepted arc})$

This is one of the most common secant-angle formulas. ---

Interior Intersection of Secant/Chord Lines

📐 Angle Formed Inside the Circle

If two chords or secant segments intersect inside the circle, then the angle formed equals half the sum of the intercepted arcs.

$\qquad \angle = \dfrac{1}{2}(\text{arc}_1 + \text{arc}_2)$

💡 Memory Rule

outside the circle: half difference

inside the circle: half sum

---

Secant-Secant Length Relation

📐 Secant-Secant Theorem

If two secants are drawn from the same external point $P$ and meet the circle at $A,B$ and $C,D$ , with $A,C$ nearer to $P$ , then

$\qquad PA\cdot PB = PC\cdot PD$

This is the secant version of power of a point. ---

Minimal Worked Examples

Example 1 Two secants from an external point form an angle. The larger intercepted arc is

140^\circ

and the smaller intercepted arc is

60^\circ

. Find the angle. Using the exterior secant-angle formula,

\qquad \angle = \dfrac{1}{2}(140^\circ - 60^\circ)

\qquad = \dfrac{1}{2}(80^\circ)=40^\circ

So the angle is

\boxed{40^\circ}

. --- Example 2 From an external point

P

, two secants meet a circle at

A,B

and

C,D

. If

\qquad PA=3,\ PB=12,\ PC=4

, find

PD

. Using secant-secant relation,

\qquad PA\cdot PB = PC\cdot PD

\qquad 3\cdot 12 = 4\cdot PD

\qquad 36 = 4PD

\qquad PD=9

Hence the answer is

\boxed{9}

. ---

How to Read Secant Diagrams

💡 Diagram Strategy

Identify whether the vertex of the angle is inside or outside the circle.

Decide whether the formula needs half-sum or half-difference.

For lengths, check whether the secants start from the same external point.

Label the nearer intersection points carefully.

---

Common Patterns

📐 Patterns to Recognize

external angle formed by two secants,

interior angle formed by intersecting chord/secant lines,

secant-secant product from one external point,

mixed arc-angle equations in circle diagrams.

---

Common Mistakes

⚠️ Avoid These Errors

❌ using half-sum for an exterior secant angle,

✅ outside means half-difference

❌ using only the internal piece of a secant in length relations,

✅ use near segment times whole secant

❌ confusing chord with secant,

✅ the secant is the whole line

❌ subtracting arcs in the wrong order,

✅ larger arc minus smaller arc

---

CMI Strategy

💡 Solve Smart

First locate the vertex of the angle.

Decide whether the angle is formed inside or outside the circle.

Use arc language before substituting numbers.

If a common external point appears, check for a secant product relation.

In harder problems, combine angle and length facts in the same figure.

---

Practice Questions

:::question type="MCQ" question="An angle formed by two secants from an external point intercepts arcs of measures

120^\circ

and

40^\circ

. The angle equals" options=["

40^\circ

","

60^\circ

","

80^\circ

","

20^\circ

"] answer="A" hint="Use half the difference of the arcs." solution="For two secants meeting outside a circle,

\qquad \angle = \dfrac{1}{2}(120^\circ-40^\circ)=\dfrac{1}{2}(80^\circ)=40^\circ

. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Two secants from the same external point satisfy

PA=2

PB=10

, and

PC=4

. Find

PD

." answer="5" hint="Use the secant-secant product relation." solution="By the secant-secant theorem,

\qquad PA\cdot PB = PC\cdot PD

\qquad 2\cdot 10 = 4\cdot PD

\qquad 20 = 4PD

\qquad PD = 5

Hence the answer is

\boxed{5}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["A secant meets a circle at two points","For an exterior angle formed by two secants, the angle equals half the sum of intercepted arcs","For an angle formed inside a circle by intersecting chords, the angle equals half the sum of intercepted arcs","Two secants from the same external point satisfy a product relation"] answer="A,C,D" hint="Separate the outside-angle and inside-angle cases." solution="1. True. This is the definition of a secant.

False. Outside the circle, the angle is half the difference, not half the sum.

True. Inside the circle, the angle is half the sum of intercepted arcs.

True. This is the secant-secant theorem.

Hence the correct answer is

\boxed{A,C,D}

." ::: :::question type="SUB" question="Two secants drawn from an external point form an angle of

35^\circ

. If the smaller intercepted arc is

50^\circ

, find the larger intercepted arc." answer="

120^\circ

" hint="Use half-difference of arcs." solution="Let the larger intercepted arc be

x^\circ

. For two secants from an external point,

\qquad 35^\circ = \dfrac{1}{2}(x - 50^\circ)

\qquad 70^\circ = x - 50^\circ

Hence

\qquad x = 120^\circ

Therefore the larger intercepted arc is

\boxed{120^\circ}

." ::: ---

Summary

❗ Key Takeaways for CMI

A secant meets a circle at two points.

Exterior angle from two secants equals half the difference of intercepted arcs.

Interior angle from intersecting chord/secant lines equals half the sum of intercepted arcs.

Two secants from the same external point satisfy a product relation.

In secant problems, the location of the vertex decides the correct formula.

---

💡 Next Up

Proceeding to Power of a point.

---

Part 5: Power of a point

Power of a Point

Overview

The power of a point is one of the most useful structural ideas in circle geometry. It connects secants, chords, and tangents through one invariant product. In exam problems, it often converts a geometry figure into a clean algebraic equation. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define the power of a point with respect to a circle,

use chord-chord, secant-secant, and tangent-secant relations,

identify when two products must be equal,

solve length problems in circles quickly,

connect the theorem with radical-axis style reasoning in harder problems.

---

Core Idea

📖 Power of a Point

Fix a circle and a point $P$ . The power of $P$ with respect to the circle is the same value obtained from any secant or tangent drawn from $P$ .

This value is:

positive if $P$ is outside the circle,

zero if $P$ lies on the circle,

negative if $P$ is inside the circle.

---

Main Relations

📐 Chord-Chord Form

If two chords $AB$ and $CD$ intersect at an interior point $P$ , then

$\qquad PA \cdot PB = PC \cdot PD$

📐 Secant-Secant Form

If two secants through an external point $P$ meet the circle at $A,B$ and $C,D$ respectively, with $A,C$ nearer to $P$ , then

$\qquad PA \cdot PB = PC \cdot PD$

📐 Tangent-Secant Form

If from an external point $P$ , a tangent $PT$ and a secant through $A,B$ are drawn, then

$\qquad PT^2 = PA \cdot PB$

❗ One Theorem, Many Shapes

These are different geometric configurations of the same power-of-a-point idea.

---

Sign and Position Interpretation

💡 How to Think About the Point

If $P$ is outside the circle, segment lengths are ordinary positive lengths and the formulas look straightforward.

If $P$ is inside the circle, the chord-chord form is the most natural statement.

If the point is on the circle, the power is zero.

---

Why the Product Stays the Same

❗ Invariance

If a point is fixed and you draw different secants through it, the product

$\qquad (\text{near segment}) \times (\text{whole secant})$

remains constant.

That is the real content of the theorem.

---

Common Configurations

📐 Most Used Setups

two chords intersect inside a circle:

\qquad PA \cdot PB = PC \cdot PD

two secants from an external point:

\qquad PA \cdot PB = PC \cdot PD

tangent and secant from an external point:

\qquad PT^2 = PA \cdot PB

---

Minimal Worked Examples

Example 1 Two chords intersect at

P

inside a circle. If

\qquad PA=2,\ PB=6,\ PC=3

, find

PD

. Using power of a point,

\qquad PA \cdot PB = PC \cdot PD

\qquad 2\cdot 6 = 3\cdot PD

\qquad 12 = 3PD

\qquad PD = 4

So the answer is

\boxed{4}

. --- Example 2 From an external point

P

, a tangent

PT

and a secant through

A,B

are drawn. If

\qquad PA=4,\ PB=9

, find

PT

. Using tangent-secant relation,

\qquad PT^2 = PA \cdot PB = 4\cdot 9 = 36

\qquad PT=6

Hence the tangent length is

\boxed{6}

. ---

How to Spot the Theorem Quickly

💡 Exam Recognition

Use power of a point whenever:

there is one special point with two lines through the circle,

segment products appear naturally,

a tangent and secant come from the same external point,

intersecting chords create four labelled pieces.

---

Common Mistakes

⚠️ Avoid These Errors

❌ multiplying the wrong pair of segments,

✅ use segments on the same line from the common point

❌ using only the internal part of a secant from an external point,

✅ use near segment times whole secant

❌ forgetting that tangent length is squared,

✅

PT^2 = PA \cdot PB

❌ mixing configurations from different points,

✅ the equality comes from one fixed point only

---

CMI Strategy

💡 Solve Smart

First identify the special point.

Check whether the point is inside or outside the circle.

Write the correct product relation immediately.

Only then substitute numbers.

In harder questions, combine power of a point with similar triangles or angle facts.

---

Practice Questions

:::question type="MCQ" question="If two chords

AB

and

CD

intersect at an interior point

P

of a circle, then which relation is correct?" options=["

PA+PB=PC+PD

","

PA\cdot PB=PC\cdot PD

","

PA\cdot PC=PB\cdot PD

","

PA+PD=PB+PC

"] answer="B" hint="Use the standard intersecting chords theorem." solution="For two chords intersecting at an interior point of a circle, the power-of-a-point relation is

\qquad PA\cdot PB = PC\cdot PD

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Two chords intersect at a point

P

inside a circle. If

PA=3

PB=8

, and

PC=4

, find

PD

." answer="6" hint="Use the chord-chord product relation." solution="By power of a point,

\qquad PA\cdot PB = PC\cdot PD

\qquad 3\cdot 8 = 4\cdot PD

\qquad 24 = 4PD

\qquad PD = 6

Hence the answer is

\boxed{6}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If a tangent

PT

and a secant through

A,B

are drawn from an external point

P

, then

PT^2=PA\cdot PB

","For two secants from the same external point, the products of near segment and whole secant are equal","The power of a point depends on which secant is chosen through the point","If a point lies on the circle, its power is

0

"] answer="A,B,D" hint="Power of a point is invariant for a fixed point." solution="1. True. This is the tangent-secant form.

True. This is the secant-secant form.

False. For a fixed point, the power is invariant.

True. A point on the circle has power

0

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="From an external point

P

, a tangent

PT

and a secant through

A,B

are drawn to a circle. If

PA=5

and

PB=20

, find

PT

." answer="

10

" hint="Use the tangent-secant theorem." solution="By power of a point,

\qquad PT^2 = PA\cdot PB

\qquad PT^2 = 5\cdot 20 = 100

Hence

\qquad PT = 10

Therefore the answer is

\boxed{10}

." ::: ---

Summary

❗ Key Takeaways for CMI

Power of a point turns circle geometry into invariant segment products.

For intersecting chords: $\qquad PA\cdot PB = PC\cdot PD$

For two secants from an external point: $\qquad PA\cdot PB = PC\cdot PD$

For tangent and secant: $\qquad PT^2 = PA\cdot PB$

The theorem is strongest when you first identify the common point correctly.

---

💡 Next Up

Proceeding to Circle-based similarity problems.

---

Part 6: Circle-based similarity problems

Circle-Based Similarity Problems

Overview

Many geometry problems involving circles are solved by finding similar triangles. The circle itself usually provides the equal angles needed for similarity through results about angles in the same segment, equal angles subtended by the same chord, and the tangent-chord theorem. In exam questions, success depends on spotting the right pair of triangles quickly. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

identify similar triangles in circle configurations,

use chord-angle and tangent-angle theorems to create equal angles,

apply AA similarity in cyclic and tangent-based figures,

convert similarity into length relations,

solve geometry problems involving secants, tangents, and chords.

---

Core Idea

📖 Similarity in Circle Geometry

In circle problems, triangles become similar when the circle gives angle equalities. The most common sources are:

angles in the same segment,

angles subtended by the same chord,

angle between tangent and chord,

cyclic quadrilateral angle relations.

---

Most Important Angle Sources

📐 Angles in the Same Segment

If points $P$ and $Q$ lie on the same segment of chord $AB$ , then

$\qquad \angle APB=\angle AQB$

📐 Tangent-Chord Theorem

The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.

If a tangent at $A$ meets chord $AB$ , then the angle between the tangent and chord $AB$ equals the angle subtended by chord $AB$ at any point on the opposite arc.

📐 Cyclic Quadrilateral Fact

If $A,B,C,D$ lie on a circle, then opposite angles satisfy

$\qquad \angle A+\angle C=180^\circ$
and
$\qquad \angle B+\angle D=180^\circ$

These angle tools often create AA similarity immediately. ---

Standard Similarity Pattern 1

📐 Chord-Based Similarity

If two triangles stand on the same chord or equal chords, then equal subtended angles often appear.

This is especially useful when triangles share a point outside or inside the circle.

---

Standard Similarity Pattern 2

📐 Tangent and Chord

Suppose $AT$ is tangent at $A$ , and $AB$ is a chord. Then the angle

$\qquad \angle TAB$

equals the angle in the alternate segment subtended by chord $AB$ .

This often gives one angle, and another angle usually comes from a shared angle or another chord theorem, producing similarity.

---

Standard Similarity Pattern 3

📐 Intersecting Chords and Similar Triangles

When two chords intersect inside a circle, the small triangles formed often become similar because of vertically opposite angles and equal subtended angles.

This type of setup leads to useful relations such as products of segments. ---

Minimal Worked Examples

Example 1 Let

AT

be tangent to a circle at

A

, and let

AB

be a chord. Let

C

be a point on the circle on the opposite arc of

AB

. Then by the tangent-chord theorem,

\qquad \angle TAB=\angle ACB

This angle equality is often enough to start a similarity argument. --- Example 2 If

A,B,C,D

are concyclic, then

\qquad \angle ABC=\angle ADC

because both subtend chord

AC

. Similarly,

\qquad \angle BAC=\angle BDC

So triangles involving these angles often become similar by AA. ---

How to Spot Similarity Fast

💡 Triangle-Hunting Method

Mark every angle coming from the same chord.

Mark tangent-chord equal angles separately.

Look for a shared angle or vertically opposite angle.

Once two angles match, use AA similarity immediately.

Then convert similarity into side ratios.

---

What Similarity Gives You

📐 After Similarity

If two triangles are similar, you can get:

ratios of corresponding sides,

equal ratios leading to unknown lengths,

proportional segments on secants and tangents,

area relations,

further angle equalities.

In many circle problems, the angle work is short, but the length work after similarity is the real target. ---

Common Patterns

📐 Patterns to Recognize

tangent plus chord plus triangle on the circle,

cyclic quadrilateral creating equal angles,

intersecting chords creating small similar triangles,

secant-tangent configurations,

two triangles subtending the same chord.

---

Common Mistakes

⚠️ Avoid These Errors

❌ using equal angles without naming the common chord,

✅ always state which chord or tangent theorem gives the angle

❌ assuming two angles are equal just because they look similar,

✅ justify each angle from circle geometry

❌ mixing up corresponding sides after similarity,

✅ write the triangle order carefully

❌ forgetting that tangent is perpendicular only to the radius at the point of contact,

✅ do not overuse the tangent property

---

CMI Strategy

💡 How to Attack Circle Similarity Problems

First look for a tangent.

If no tangent, look for equal angles subtended by a common chord.

Mark all cyclic angle equalities.

Search for two triangles with two equal angles.

Once similarity is found, switch from angle language to ratio language immediately.

---

Practice Questions

:::question type="MCQ" question="Which theorem is most commonly used to create similarity when a tangent and a chord appear together in a circle problem?" options=["Pythagoras theorem","Tangent-chord theorem","Angle bisector theorem","Ceva's theorem"] answer="B" hint="Think about the angle between tangent and chord." solution="When a tangent and a chord appear together, the most useful angle relation is given by the tangent-chord theorem. This frequently produces one of the equal angles needed for similarity. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="If two triangles in a circle configuration have two equal angles, how many degrees is the third pair of corresponding angles apart?" answer="0" hint="Use the angle sum of a triangle." solution="If two angles of one triangle equal two angles of another triangle, then the third angles are also equal because each triangle has angle sum

180^\circ

. Therefore the difference between the third pair is

\boxed{0}

degrees." ::: :::question type="MSQ" question="Which of the following are standard sources of similarity in circle geometry?" options=["Angles in the same segment","Tangent-chord theorem","Cyclic quadrilateral angle relations","The fact that all radii are parallel"] answer="A,B,C" hint="Look for angle-producing theorems." solution="1. True. Angles in the same segment often give equal angles.

True. The tangent-chord theorem is one of the main sources of angle equality.

True. Cyclic quadrilateral angle relations also produce equal or supplementary angles useful for similarity.

False. Radii are not all parallel.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Explain why circle-based similarity problems are usually solved by angle equalities before side ratios." answer="Because the circle primarily provides angle equalities, and similarity follows from AA before any side relation can be used" hint="Think about what theorems a circle gives most naturally." solution="In most circle problems, the main theorems give angle information: angles in the same segment, equal angles subtended by the same chord, tangent-chord theorem, and cyclic angle relations. These allow us to show that two triangles are similar by AA. Once similarity is established, we then obtain side ratios and length relations. So angle equalities usually come first, and side ratios come after similarity is proved." ::: ---

Summary

❗ Key Takeaways for CMI

Circle similarity problems are mainly angle-chasing problems.

Same-chord angles and tangent-chord angles are the main tools.

The goal is usually to prove AA similarity quickly.

After similarity, convert to side ratios immediately.

Precise theorem justification matters in geometry proofs.

---

Chapter Summary

❗ Circles — Key Points

Angle Theorems: Mastery of angles subtended by arcs, inscribed angles, angles in a semi-circle, and properties of cyclic quadrilaterals is fundamental for solving angle-related problems.
Chord Properties: Understanding the relationship between chords, their distance from the center, and the perpendicular bisector theorem is crucial for calculating lengths and distances.
Tangents and Secants: Key theorems include the tangent-radius perpendicularity, properties of two tangents from an external point, and the tangent-secant theorem for segment lengths.
Power of a Point: This unifying concept describes the product of segment lengths for intersecting chords, secants, and tangents from an external point, often derived through similar triangles.
* Circle-based Similarity: Recognizing similar triangles formed by chords, secants, and tangents (e.g., in Power of a Point scenarios) is vital for solving complex geometric problems and deriving relations.

---

Chapter Review Questions

:::question type="MCQ" question="A cyclic quadrilateral ABCD has $\angle A = 75^\circ$ and $\angle B = 85^\circ$ . What is the measure of $\angle C$ ?" options=[" $105^\circ$ "," $95^\circ$ "," $75^\circ$ "," $85^\circ$ "] answer=" $105^\circ$ " hint="Opposite angles of a cyclic quadrilateral are supplementary." solution="For a cyclic quadrilateral, $\angle A + \angle C = 180^\circ$ . Given $\angle A = 75^\circ$ , we have $\angle C = 180^\circ - 75^\circ = 105^\circ$ ."
:::

:::question type="NAT" question="From an external point P, a tangent PT and a secant PAB are drawn to a circle. If PT = 8 and PA = 4, find the length of PB." answer="16" hint="Apply the Tangent-Secant Theorem (Power of a Point)." solution="According to the Tangent-Secant Theorem, $PT^2 = PA \cdot PB$ . Substituting the given values: $8^2 = 4 \cdot PB \implies 64 = 4 \cdot PB \implies PB = \frac{64}{4} = 16$ ."
:::

:::question type="MCQ" question="A circle has a radius of 13 cm. A chord of length 24 cm is drawn in the circle. What is the distance of the chord from the center of the circle?" options=["5 cm","10 cm","12 cm","13 cm"] answer="5 cm" hint="The perpendicular from the center to a chord bisects the chord. Use the Pythagorean theorem." solution="Let the radius be $r=13$ cm and the chord length be $L=24$ cm. The perpendicular from the center bisects the chord, so half the chord length is $L/2 = 12$ cm. Let the distance from the center be $d$ . By the Pythagorean theorem, $d^2 + (L/2)^2 = r^2$ . So, $d^2 + 12^2 = 13^2 \implies d^2 + 144 = 169 \implies d^2 = 169 - 144 = 25 \implies d = 5$ cm."
:::

:::question type="NAT" question="Chords AB and CD of a circle intersect at point E. If AE = 9, EB = 4, and CE = 6, find the length of ED." answer="6" hint="Apply the Intersecting Chords Theorem (Power of a Point)." solution="According to the Intersecting Chords Theorem, $AE \cdot EB = CE \cdot ED$ . Substituting the given values: $9 \cdot 4 = 6 \cdot ED \implies 36 = 6 \cdot ED \implies ED = \frac{36}{6} = 6$ ."
:::

---

What's Next?

💡 Continue Your CMI Journey

This foundational understanding of circles is crucial for advanced topics in Euclidean geometry, including the study of triangle centers (e.g., circumcenter, incenter) and their associated circles. The principles of similarity and congruence, extensively used in circle theorems, will continue to be vital in the study of other geometric figures like triangles and quadrilaterals, as well as in applying geometric reasoning within coordinate systems.