I'm Up and Down, and Round and Round — Notes

Definitions

A circle is defined as the set of all points in a plane that are the same distance from a fixed point in that plane. This fixed point is called the centre of the circle.

• The "set of all points" means we are considering every single point that meets the condition.
• The "plane" refers to a flat, two-dimensional surface.
• The "same distance" is a constant value, known as the radius.
• The "fixed point" is the centre of the circle.

This set of points that satisfy a specific condition is also known as the locus of points. So, a circle is the locus of points equidistant from a given point.

In Figure 5.3:

• A is the centre of the circle.
• The radius of the circle is the distance from the centre to any point on the circle. If B is a point on the circle, then the length of the line segment AB is the radius.

A line segment connecting any two points on the circle is called a chord. In Figure 5.3, BC is a chord.

The angle subtended by a chord at the centre is the angle formed by the two radii connecting the endpoints of the chord to the centre. In Figure 5.3, the angle subtended by chord BC at centre A is $\mathrm{\angle }BAC$.

A diameter is a chord that passes through the centre of the circle. A diameter is also twice the length of the radius ($d=2r$).

Example 1: Identifying Circle Components

Given: A circle with centre O and a point P on the circle. A line segment QR where Q and R are points on the circle, and O lies on QR. To find: Identify the radius, a chord, and the diameter.

Solution:

1. The distance from centre O to point P is the radius.
2. The line segment QR connects two points on the circle. Since O lies on QR, QR is a diameter.
3. A chord is any line segment connecting two points on the circle. So, QR is also a chord, and any other line segment connecting two points on the circle (like QS, if S is another point on the circle) would also be a chord.

Answer: Radius: OP, Chord: QR (and others), Diameter: QR.

Example 2: Calculating Radius and Diameter

Given: A circle with centre C and radius $r=5$ cm. A chord AB passes through C. To find: The length of the diameter.

Solution:

1. A chord passing through the centre is a diameter.
2. The diameter ($d$) is twice the radius ($r$).
3. $d=2\times r$
4. $d=2\times 5$ cm
5. $d=10$ cm

Answer: The diameter is 10 cm.

Example 3: Understanding Locus

Given: A point P is always at a distance of 7 units from the origin (0,0) in a coordinate plane. To find: Describe the locus of point P.

Solution:

1. The condition given is that the distance of point P from the fixed point (0,0) is constant (7 units).
2. This is the definition of a circle.
3. The fixed point (0,0) is the centre.
4. The constant distance (7 units) is the radius.
5. Therefore, the locus of point P is a circle with centre at the origin and radius 7 units.

Answer: The locus of point P is a circle with centre (0,0) and radius 7 units.

Remember

A circle is a 2D shape. All points on the circle are equidistant from the centre.

Exam-smart

You might be given a description like "the set of points 3 cm away from point A". You need to recognise this as a circle and identify its centre and radius. Or you might be given a diagram with a circle and asked to label the centre, radius, chord, and diameter.

Symmetries of a Circle

A circle is defined as the set of all points in a plane that are equidistant from a fixed point. This fixed point is called the centre, and the constant distance is called the radius. The unique properties of a circle stem from its perfect symmetry.

Rotational Symmetry

A circle possesses complete rotational symmetry. This means that if you rotate a circle about its centre by any angle, it will look exactly the same. Imagine a wheel on a vehicle; you cannot distinguish between different points of the wheel touching the ground after rotation because the wheel appears identical from every angle.

This complete rotational symmetry is a fundamental characteristic that distinguishes circles from other shapes like squares or regular polygons, which have rotational symmetry only for specific angles.

Reflectional Symmetry

A circle has reflectional symmetry along any line that passes through its centre. Such a line is called a diameter. If you fold a circular cutout along any of its diameters, the two halves will perfectly overlap, demonstrating that each diameter is an axis of symmetry.

• A line of symmetry (or axis of symmetry) is a line that divides a shape into two identical halves, such that one half is the mirror image of the other.
• A diameter of a circle is a line segment that passes through the centre and has its endpoints on the circle. It is also the longest chord of a circle.

Example: Lines of Symmetry

Given: A circle with centre O.

To find: The lines of reflectional symmetry of the circle.

Solution:

1. Consider any line passing through the centre O.
2. Let this line intersect the circle at points P and Q. The line segment PQ is a diameter.
3. If we fold the circle along the line PQ, the two semicircles formed will coincide exactly.
4. This means the line PQ is a line of symmetry.
5. Since we can draw infinitely many such lines passing through the centre, a circle has infinitely many lines of symmetry.

Answer: A circle has infinitely many lines of symmetry, all of which are its diameters.

Comparison: Symmetries of a Circle vs. a Square

• Thinking about the "Why": The circle's symmetry is due to its definition. Every point on the circle is equidistant from the center. This uniform distance ensures that no matter how you rotate or reflect it through its center, the shape remains indistinguishable. A square, while symmetrical, has defined vertices and sides, limiting its rotational and reflectional symmetry to specific orientations.

Remember

A circle's defining characteristic—its uniform radius—is the source of its perfect rotational and reflectional symmetry. Any line through the center is an axis of symmetry.

Exam-smart

You might be given a diagram of a circle with several lines drawn through it, some passing through the centre and some not. You'd be asked to identify which of these lines are axes of symmetry. The key is that only lines passing through the centre are lines of symmetry for a circle.

How Many Circles?

This section explores how many circles can be drawn through a given number of points on a plane.

Circles Through Two Points

Given two distinct points $A$ and $B$ on a plane, an infinite number of circles can pass through both of them.

• Why? For a circle to pass through $A$ and $B$, its centre must be equidistant from $A$ and $B$.
• The locus of points equidistant from two given points is the perpendicular bisector of the line segment joining those two points.
• Therefore, any point on the perpendicular bisector of segment $AB$ can serve as the centre of a circle passing through $A$ and $B$.

Properties of Circles Through Two Points

Example 1

Given: Two points $P$ and $Q$ on a plane.

To find: How many circles can pass through $P$ and $Q$?

Solution: The centre of any circle passing through $P$ and $Q$ must be equidistant from $P$ and $Q$. The locus of points equidistant from $P$ and $Q$ is the perpendicular bisector of segment $PQ$. Since there are infinitely many points on the perpendicular bisector, there can be infinitely many centres. Each centre, with a suitable radius, defines a unique circle passing through $P$ and $Q$. Therefore, infinitely many circles can pass through two given points $P$ and $Q$.

Answer: Infinitely many circles.

Try this:

What is the radius of the smallest circle that passes through two points that are 10 cm apart? Answer: 5 cm

Circles Through Three Points

• Collinear Points: If three points $A$, $B$, and $C$ lie on a straight line (collinear), then no circle can pass through all three points. This is because the perpendicular bisectors of segments $AB$ and $BC$ would be parallel (or the same line if $A$, $B$, $C$ are equally spaced, which is a special case of parallel), and parallel lines do not intersect at a unique point to form a centre. A circle cannot intersect a line at three distinct points.

• Non-collinear Points: If three points $A$, $B$, and $C$ do not lie on a straight line (non-collinear), then there is exactly one unique circle that passes through all three points.

• Why? For a circle to pass through non-collinear points $A$, $B$, and $C$, its centre $O$ must be equidistant from all three points ($OA=OB=OC$).

• Since $OA=OB$, $O$ must lie on the perpendicular bisector of $AB$.

• Since $OB=OC$ (or $OA=OC$), $O$ must also lie on the perpendicular bisector of $BC$ (or $AC$).

• As $A$, $B$, and $C$ are non-collinear, the perpendicular bisectors of $AB$ and $BC$ are not parallel and will intersect at exactly one point. This unique intersection point is the centre $O$.

• The radius of this unique circle is the distance from $O$ to any of the points $A$, $B$, or $C$.

The Circumcircle and Circumcentre

• When three non-collinear points form a triangle ($\mathrm{△}ABC$), the unique circle passing through its vertices ($A$, $B$, and $C$) is called the circumcircle.
• The centre of this circumcircle is called the circumcentre of the triangle. It is the point where the perpendicular bisectors of the sides of the triangle intersect.
• The triangle is said to be inscribed in the circumcircle, and the circumcircle circumscribes the triangle.

Position of the Circumcentre

The location of the circumcentre relative to the triangle depends on the type of triangle:

• Acute-angled triangle: The circumcentre lies inside the triangle.
• Obtuse-angled triangle: The circumcentre lies outside the triangle.
• Right-angled triangle: The circumcentre lies at the midpoint of the hypotenuse.

Example 2

Given: A triangle $\mathrm{△}ABC$ with vertices $A=(0,0)$, $B=(4,0)$, and $C=(2,3)$.

To find: The circumcircle passing through $A,B,C$.

Solution:

1. Find the midpoint of $AB$: $M_{AB}=(\frac{0+4}{2},\frac{0+0}{2})=(2,0)$. The slope of $AB$ is $0$. The perpendicular bisector of $AB$ is a vertical line passing through $(2,0)$, so its equation is $x=2$.
2. Find the midpoint of $BC$: $M_{BC}=(\frac{4+2}{2},\frac{0+3}{2})=(3,\frac{3}{2})$. The slope of $BC$ is $\frac{3-0}{2-4}=\frac{3}{-2}=-\frac{3}{2}$. The slope of the perpendicular bisector of $BC$ is the negative reciprocal: $m_{\perp }=-\frac{1}{-\frac{3}{2}}=\frac{2}{3}$. The equation of the perpendicular bisector of $BC$ is $y-\frac{3}{2}=\frac{2}{3}(x-3)$.

$y=\frac{2}{3}x-\frac{1}{2}$. 3. The circumcentre $O$ is the intersection of $x=2$ and $y=\frac{2}{3}x-\frac{1}{2}$. Substitute $x=2$ into the second equation: $y=\frac{2}{3}(2)-\frac{1}{2}=\frac{4}{3}-\frac{1}{2}=\frac{8}{6}-\frac{3}{6}=\frac{5}{6}$. So, the circumcentre is $O=(2,\frac{5}{6})$. 4. The radius $R$ is the distance from $O$ to any vertex, say $A=(0,0)$. $R=\sqrt{(2-0{)}^2+{(\frac{5}{6}-0)}^2}=\sqrt{2^2+{\left(\frac{5}{6}\right)}^2}=\sqrt{4+\frac{25}{36}}=\sqrt{\frac{144}{36}+\frac{25}{36}}=\sqrt{\frac{169}{36}}=\frac{13}{6}$. 5. The equation of the circumcircle is $(x-h{)}^2+(y-k{)}^2=R^2$, where $(h,k)$ is the centre. $(x-2{)}^2+{(y-\frac{5}{6})}^2={\left(\frac{13}{6}\right)}^2=\frac{169}{36}$.

Answer: The circumcentre is $(2,\frac{5}{6})$ and the equation of the circumcircle is $(x-2{)}^2+{(y-\frac{5}{6})}^2=\frac{169}{36}$.

Try this:

How many circles can pass through three collinear points? Answer: Zero

Remember

Two points define infinitely many circles, but three non-collinear points define exactly one circle.

Exam-smart

You might be given coordinates of three points and asked to find the equation of the circle passing through them. This involves finding the perpendicular bisectors of two sides and their intersection point (the circumcentre), then calculating the distance from the centre to one of the points (the radius). Or, you might be given the circumcircle's equation and asked properties of the inscribed triangle.

Chords and the Angles They Subtend

A chord of a circle is a line segment whose endpoints both lie on the circle. Imagine a thread tied between two points on a wheel; this thread represents a chord. The angle subtended by a chord at the centre is the angle formed by joining the endpoints of the chord to the centre of the circle.

Theorem 2: Equal chords of a circle subtend equal angles at the centre.

This theorem states that if two chords in the same circle have the same length, then the angles they form at the circle's centre will also be equal.

Given: A circle with centre C, and two chords AB and DE such that AB = DE. To show: The angle subtended by AB at the centre (∠ACB) is equal to the angle subtended by DE at the centre (∠DCE).

Proof:

1. In $\mathrm{△}ACB$, $CA=CB$ (both are radii of the circle).
2. In $\mathrm{△}DCE$, $CD=CE$ (both are radii of the circle).
3. Therefore, $CA=CD$ and $CB=CE$.
4. We are given that $AB=DE$.
5. By the SSS (Side-Side-Side) congruence criterion, $\mathrm{△}CAB\cong \mathrm{△}CDE$.
6. Since the triangles are congruent, their corresponding angles are equal. Thus, $\mathrm{\angle }ACB=\mathrm{\angle }DCE$.

This means that if chords are of equal length, they will 'reach out' to the centre with the same angle.

Example 1

Given: In a circle with centre O, chord AB = chord CD. To find: If $\mathrm{\angle }AOB={70}^{\circ }$, find $\mathrm{\angle }COD$.

Solution:

1. The theorem states that equal chords subtend equal angles at the centre.
2. We are given that chord AB = chord CD.
3. Therefore, the angle subtended by AB at the centre, $\mathrm{\angle }AOB$, must be equal to the angle subtended by CD at the centre, $\mathrm{\angle }COD$.
4. Given $\mathrm{\angle }AOB={70}^{\circ }$.
5. So, $\mathrm{\angle }COD=\mathrm{\angle }AOB$.
6. $\mathrm{\angle }COD={70}^{\circ }$.

Answer: $\mathrm{\angle }COD={70}^{\circ }$.

Theorem 3: Chords of a circle that subtend equal angles at the centre are equal.

This is the converse of Theorem 2. It states that if two chords in the same circle form equal angles at the centre, then the chords themselves must have the same length.

Given: A circle with centre C, and two chords AB and DE such that $\mathrm{\angle }ACB=\mathrm{\angle }DCE$. To show: The length of chord AB is equal to the length of chord DE (AB = DE).

Proof:

1. In $\mathrm{△}ACB$, $AC=BC$ (both are radii of the circle).
2. In $\mathrm{△}DCE$, $DC=EC$ (both are radii of the circle).
3. Therefore, $AC=DC$ and $BC=EC$.
4. We are given that $\mathrm{\angle }ACB=\mathrm{\angle }DCE$.
5. By the SAS (Side-Angle-Side) congruence criterion, $\mathrm{△}ACB\cong \mathrm{△}DCE$.
6. Since the triangles are congruent, their corresponding sides are equal. Thus, $AB=DE$.

This means that if chords create the same angle at the centre, they must be of the same length.

Example 2

Given: In a circle with centre O, $\mathrm{\angle }POQ=\mathrm{\angle }ROS$. Chord PQ has length 8 cm. To find: The length of chord RS.

Solution:

1. The theorem states that chords subtending equal angles at the centre are equal.
2. We are given that $\mathrm{\angle }POQ=\mathrm{\angle }ROS$.
3. Therefore, the chord PQ must be equal in length to the chord RS.
4. Given chord PQ = 8 cm.
5. So, chord RS = chord PQ.
6. Chord RS = 8 cm.

Answer: Chord RS = 8 cm.

Example 3

Given: In the figure, AB and CD are chords of a circle with centre O. If $\mathrm{\angle }AOB=\mathrm{\angle }COD$, prove that AB = CD.

To prove: AB = CD.

Proof:

1. Consider $\mathrm{△}AOB$ and $\mathrm{△}COD$.
2. $OA=OB$ (radii of the same circle).
3. $OC=OD$ (radii of the same circle).
4. Thus, $OA=OC$ and $OB=OD$.
5. We are given that $\mathrm{\angle }AOB=\mathrm{\angle }COD$.
6. By the SAS congruence criterion, $\mathrm{△}AOB\cong \mathrm{△}COD$.
7. Since the triangles are congruent, their corresponding sides are equal.
8. Therefore, AB = CD.

Answer: AB = CD is proved.

Remember

The relationship between chord length and the angle it subtends at the centre is a direct one: equal chords mean equal angles, and equal angles mean equal chords.

Exam-smart

In a source-based question, you might be given a diagram of a circle with two chords and their lengths or the angles they subtend at the centre. You would apply these theorems to find unknown lengths or angles, or to prove relationships between chords. For example, if you are shown two chords with the same length, you know they subtend the same angle at the centre, even if the angle value is not given. Conversely, if two angles at the centre are shown to be equal, you can conclude the chords are of equal length.

Midpoints and Perpendicular Bisectors of Chords

This section explores the relationship between the centre of a circle, the midpoint of a chord, and the perpendicular bisector of a chord. These properties are crucial for understanding the geometry of circles and solving problems involving chords.

Theorem 4: The Line Joining the Centre to the Midpoint of a Chord is Perpendicular to the Chord

Statement: If a line segment is drawn from the centre of a circle to the midpoint of a chord, then this line segment is perpendicular to the chord.

Proof: Consider a circle with centre C and a chord AB. Let M be the midpoint of AB. This means $AM=MB$. We need to show that CM is perpendicular to AB, i.e., $\mathrm{\angle }CMA=\mathrm{\angle }CMB={90}^{\circ }$.

1. Consider $\mathrm{△}CMA$ and $\mathrm{△}CMB$:

• $CA=CB$ (radii of the same circle).
• $AM=MB$ (given that M is the midpoint of AB).
• $CM=CM$ (common side).

2. Congruence: By the SSS (Side-Side-Side) congruence criterion, $\mathrm{△}CMA\cong \mathrm{△}CMB$.

3. Corresponding Angles: Since the triangles are congruent, their corresponding angles are equal. Therefore, $\mathrm{\angle }CMA=\mathrm{\angle }CMB$.

4. Angles on a Straight Line: Angles $\mathrm{\angle }CMA$ and $\mathrm{\angle }CMB$ form a linear pair on the straight line AB. Thus, $\mathrm{\angle }CMA+\mathrm{\angle }CMB={180}^{\circ }$.

5. Conclusion: Substituting $\mathrm{\angle }CMA=\mathrm{\angle }CMB$ into the linear pair equation, we get $2\mathrm{\angle }CMA={180}^{\circ }$. This implies $\mathrm{\angle }CMA={90}^{\circ }$. Since $\mathrm{\angle }CMA={90}^{\circ }$, then $\mathrm{\angle }CMB$ is also ${90}^{\circ }$. Therefore, CM is perpendicular to AB.

Example 1

Given: A circle with centre O. A chord AB and M is the midpoint of AB. To find: Prove that $OM\perp AB$.

Solution:

1. Consider $\mathrm{△}OMA$ and $\mathrm{△}OMB$.
2. $OA=OB$ (radii of the same circle).
3. $AM=MB$ (M is the midpoint of AB).
4. $OM=OM$ (common side).
5. By SSS congruence, $\mathrm{△}OMA\cong \mathrm{△}OMB$.
6. Therefore, $\mathrm{\angle }OMA=\mathrm{\angle }OMB$.
7. Also, $\mathrm{\angle }OMA+\mathrm{\angle }OMB={180}^{\circ }$ (linear pair).
8. So, $2\mathrm{\angle }OMA={180}^{\circ }$.
9. $\mathrm{\angle }OMA={90}^{\circ }$.
10. Thus, $OM\perp AB$. Answer: Proof completed.

Theorem 5: The Perpendicular from the Centre to a Chord Bisects the Chord

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

Proof: Consider a circle with centre C and a chord AB. Draw a perpendicular from C to AB, and let the point of intersection be M. So, $\mathrm{\angle }CMA={90}^{\circ }$. We need to show that M is the midpoint of AB, i.e., $AM=MB$.

1. Consider $\mathrm{△}CMA$ and $\mathrm{△}CMB$:

• $CA=CB$ (radii of the same circle).
• $\mathrm{\angle }CMA=\mathrm{\angle }CMB={90}^{\circ }$ (given that CM is perpendicular to AB).
• $CM=CM$ (common side).

2. Congruence: By the RHS (Right angle-Hypotenuse-Side) congruence criterion, $\mathrm{△}CMA\cong \mathrm{△}CMB$.

3. Corresponding Sides: Since the triangles are congruent, their corresponding sides are equal. Therefore, $AM=MB$.

4. Conclusion: Thus, the perpendicular from the centre C to the chord AB bisects the chord AB at point M.

Example 2

Given: In a circle with centre O, a chord AB has length 16 cm. The distance of the chord from the centre is 6 cm. To find: The radius of the circle.

Solution:

1. Draw a perpendicular from O to AB, meeting AB at M.
2. Since the perpendicular from the centre bisects the chord, $AM=MB=\frac{1}{2}AB$.
3. $AM=\frac{1}{2}\times 16\text{ cm}=8\text{ cm}$.
4. We are given $OM=6\text{ cm}$.
5. Consider the right-angled triangle $\mathrm{△}OMA$. By the Pythagorean theorem:

6. $O{A}^2=(6\text{ cm}{)}^2+(8\text{ cm}{)}^2$
7. $O{A}^2=36{\text{ cm}}^2+64{\text{ cm}}^2$
8. $O{A}^2=100{\text{ cm}}^2$
9. $OA=\sqrt{100{\text{ cm}}^2}=10\text{ cm}$.
10. OA is the radius of the circle. Answer: The radius of the circle is 10 cm.

Example 3

Given: Two chords AB and CD of lengths 10 cm and 24 cm respectively, are in a circle of radius 13 cm. To find: The distance of each chord from the centre.

Solution: Let O be the centre of the circle. Let OM be the perpendicular from O to AB, and ON be the perpendicular from O to CD. Since the perpendicular from the centre bisects the chord: $AM=MB=\frac{1}{2}AB=\frac{1}{2}\times 10\text{ cm}=5\text{ cm}$. $CN=ND=\frac{1}{2}CD=\frac{1}{2}\times 24\text{ cm}=12\text{ cm}$. The radius of the circle is $OA=OB=OC=OD=13\text{ cm}$.

For chord AB:

1. Consider right-angled $\mathrm{△}OMA$.
2. By Pythagorean theorem: $O{A}^2=O{M}^2+A{M}^2$.
3. $(13\text{ cm}{)}^2=O{M}^2+(5\text{ cm}{)}^2$.
4. $169{\text{ cm}}^2=O{M}^2+25{\text{ cm}}^2$.
5. $O{M}^2=169{\text{ cm}}^2-25{\text{ cm}}^2=144{\text{ cm}}^2$.
6. $OM=\sqrt{144{\text{ cm}}^2}=12\text{ cm}$. The distance of chord AB from the centre is 12 cm.

For chord CD:

1. Consider right-angled $\mathrm{△}ONC$.
2. By Pythagorean theorem: $O{C}^2=O{N}^2+C{N}^2$.
3. $(13\text{ cm}{)}^2=O{N}^2+(12\text{ cm}{)}^2$.
4. $169{\text{ cm}}^2=O{N}^2+144{\text{ cm}}^2$.
5. $O{N}^2=169{\text{ cm}}^2-144{\text{ cm}}^2=25{\text{ cm}}^2$.
6. $ON=\sqrt{25{\text{ cm}}^2}=5\text{ cm}$. The distance of chord CD from the centre is 5 cm. Answer: The distance of chord AB from the centre is 12 cm, and the distance of chord CD from the centre is 5 cm.

The Perpendicular Bisector of a Chord

The perpendicular bisector of a chord is a line that cuts the chord into two equal halves and is at a 90-degree angle to it.

• Property 1: The perpendicular bisector of a chord passes through the centre of the circle. This is a direct consequence of Theorem 5. If a line bisects a chord and is perpendicular to it, it must pass through the centre.

• Property 2: If a line passes through the centre of a circle and bisects a chord, it is perpendicular to the chord. This is Theorem 4.

• Property 3: If a line passes through the centre of a circle and is perpendicular to a chord, it bisects the chord. This is Theorem 5.

These properties mean that the perpendicular bisector of a chord is essentially a line that contains both the perpendicular from the centre to the chord and the line segment from the centre to the midpoint of the chord.

Example 4 (Converse of Theorem 4)

Given: In a circle with centre O, a line segment OM is drawn such that M is the midpoint of chord AB, and $\mathrm{\angle }OMA={90}^{\circ }$. To find: Prove that OM passes through the centre O. (This is implicitly covered by Theorem 5, but we will show it from the given.)

Solution:

1. Given: M is the midpoint of AB, so $AM=MB$.
2. Given: $\mathrm{\angle }OMA={90}^{\circ }$. This means OM is perpendicular to AB.
3. We know from Theorem 5 that the perpendicular from the centre of a circle to a chord bisects the chord.
4. Since OM is perpendicular to chord AB and M is its midpoint, OM must be the line that passes through the centre of the circle.
5. Therefore, if OM is perpendicular to chord AB, then O must be the centre of the circle. The line OM contains the centre. Answer: Proof completed.

Example 5

Given: An isosceles triangle ABC is inscribed in a circle, with AB = AC. To find: Show that the altitude from A to BC passes through the centre of the circle.

Solution:

1. In $\mathrm{△}ABC$, $AB=AC$. This means $\mathrm{△}ABC$ is isosceles.
2. Let AD be the altitude from A to BC. So, $\mathrm{\angle }ADB={90}^{\circ }$.
3. In $\mathrm{△}ABD$ and $\mathrm{△}ACD$:

• $AB=AC$ (given).
• $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$ (AD is altitude).
• $AD=AD$ (common side).

4. By RHS congruence, $\mathrm{△}ABD\cong \mathrm{△}ACD$.
5. Therefore, $BD=CD$. This means D is the midpoint of BC.
6. So, AD is the line segment from vertex A to the midpoint D of the base BC, and AD is perpendicular to BC.
7. Consider BC as a chord of the circle. The line AD is perpendicular to the chord BC and D is its midpoint.
8. From Theorem 4, the line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.
9. From Theorem 5, the perpendicular from the centre to a chord bisects the chord.
10. Since AD is perpendicular to chord BC and bisects it (at D), the line AD must pass through the centre of the circle. Answer: Proof completed.

Remember

The line joining the centre to the midpoint of a chord is perpendicular to the chord. Conversely, the perpendicular from the centre to a chord bisects the chord. These two statements are essentially two sides of the same coin.

Exam-smart

Questions might provide a diagram of a circle with a chord, its midpoint, and the centre. You'll be asked to find a length (like radius or distance from centre) or prove an angle is ${90}^{\circ }$. You must recognize that a right-angled triangle is formed by the radius, the distance from the centre to the midpoint of the chord, and half the chord's length. Pythagorean theorem will be key.

Distance of Chords from the Centre

The distance of a chord from the centre of a circle is defined as the length of the perpendicular drawn from the centre to the chord. This perpendicular bisects the chord, a fact established in Theorem 5 from the previous section.

Theorem 6: Equal Chords are Equidistant from the Centre

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

Explanation: This theorem is based on the symmetry of a circle. If two chords have the same length, they are essentially interchangeable positions within the circle due to rotation. The perpendicular distance from the centre to each chord will remain the same because the circle itself is a symmetric figure.

We can prove this using congruent triangles:

1. Using SSS Congruence: Consider two equal chords, AB and FG, with centre C. Let E and H be the midpoints of AB and FG respectively. CE and CH are the perpendiculars from C to AB and FG.

• Given: AB = FG
• Given: CE ⊥ AB and CH ⊥ FG (by definition of distance from centre)
• From Theorem 5, E and H are midpoints, so AE = EB and FH = HG.
• Since AB = FG, then AE = EB = FH = HG = $\frac{1}{2}$AB.
• Consider ∆CAB and ∆CFG.
• CA = CF (radii)
• CB = CG (radii)
• AB = FG (given)
• By SSS congruence, ∆CAB ≅ ∆CFG.
• Since the triangles are congruent, their corresponding altitudes (which are the distances CE and CH) must be equal. Thus, CE = CH.

2. Using RHS Congruence:

• Given: AB = FG.
• Since E and H are midpoints, AE = FH = $\frac{1}{2}$AB.
• ∠CEA = ∠CHF = 90° (CE ⊥ AB, CH ⊥ FG).
• CA = CF (radii).
• Consider ∆CEA and ∆CHF.
• Hypotenuse CA = Hypotenuse CF (radii).
• Side AE = Side FH (since AB = FG).
• Right angle at E and H.
• By RHS congruence, ∆CEA ≅ ∆CHF.
• Therefore, the corresponding sides CE = CH.

Example 1:

Given: A circle with centre O. Chord AB = Chord CD. Radius OA = 10 cm. To find: The distance of chord AB from the centre O, if the distance of chord CD from the centre is 8 cm.

Solution:

• Let OE ⊥ AB and OF ⊥ CD.
• Given OE = 8 cm (distance of CD from centre).
• We are given AB = CD.
• By Theorem 6, equal chords are equidistant from the centre.
• Therefore, the distance of chord AB from the centre is equal to the distance of chord CD from the centre.
• OE = OF.
• So, the distance of chord AB from the centre is 8 cm.

Answer: 8 cm.

Theorem 7: Chords Equidistant from the Centre are Equal

Statement: Chords of a circle that are equidistant from the centre have equal length.

Explanation: This is the converse of Theorem 6. If two chords are the same distance from the centre, they must have the same length. This is also due to the symmetry of the circle.

Proof:

• Let AB and FG be two chords of a circle with centre C.
• Let CE ⊥ AB and CH ⊥ FG, where E and H are midpoints of AB and FG.
• Given: CE = CH (distances from centre are equal).
• Consider ∆CEA and ∆CHF.
• CE = CH (given).
• ∠CEA = ∠CHF = 90°.
• CA = CF (radii).
• By RHS congruence, ∆CEA ≅ ∆CHF.
• Therefore, the corresponding sides AE = FH.
• Since E and H are midpoints, AB = 2AE and FG = 2FH.
• Thus, AB = FG.

Example 2:

Given: In a circle with centre O, chord AB is at a distance of 5 cm from the centre. Chord CD is also at a distance of 5 cm from the centre. Radius of the circle is 13 cm. To find: The length of chord AB and chord CD.

Solution:

• Let OE ⊥ AB and OF ⊥ CD.
• Given OE = 5 cm and OF = 5 cm. Radius r = 13 cm.
• In right-angled triangle ∆OEA, by Pythagoras theorem:
• $O{A}^2=O{E}^2+A{E}^2$
• ${13}^2=5^2+A{E}^2$
• $169=25+A{E}^2$
• $A{E}^2=169-25$
• $A{E}^2=144$
• $AE=\sqrt{144}=12$ cm.
• Since OE ⊥ AB, E is the midpoint of AB.
• Length of chord AB = $2\times AE=2\times 12=24$ cm.
• Similarly, for chord CD:
• In right-angled triangle ∆OFC, by Pythagoras theorem:
• $O{C}^2=O{F}^2+F{C}^2$
• ${13}^2=5^2+F{C}^2$
• $FC=12$ cm.
• Length of chord CD = $2\times FC=2\times 12=24$ cm.
• Alternatively, using Theorem 7: Since the distances of AB and CD from the centre are equal (5 cm), their lengths must be equal.

Answer: Length of chord AB = 24 cm, Length of chord CD = 24 cm.

Theorem 8: Longer Chord is Closer to the Centre

Statement: Let AB and DE be two chords of a circle with centre C. If AB > DE, then the distance from C to AB is less than the distance from C to DE.

Explanation: This theorem states an inverse relationship between the length of a chord and its distance from the centre. The longer a chord is, the closer it lies to the centre of the circle. Conversely, shorter chords are farther away. The diameter is the longest possible chord and passes through the centre, thus its distance from the centre is 0. A chord of length zero can be considered a point, and its distance from the centre would be the radius.

Proof:

• Let AB and DE be two chords of a circle with centre C.
• Let CF ⊥ AB and CG ⊥ DE. So, CF and CG are the distances from the centre.
• Given: AB > DE.
• From Theorem 5, F and G are midpoints of AB and DE.
• So, AF = $\frac{1}{2}$AB and DG = $\frac{1}{2}$DE.
• Since AB > DE, it follows that AF > DG.
• Consider right-angled triangles ∆CFA and ∆CGD.
• CA = CD (radii).
• By Pythagoras theorem:
• $C{A}^2=C{F}^2+A{F}^2$
• $C{D}^2=C{G}^2+D{G}^2$
• Since CA = CD, we have $C{F}^2+A{F}^2=C{G}^2+D{G}^2$.
• Rearranging: $C{F}^2-C{G}^2=D{G}^2-A{F}^2$.
• Since AF > DG, $A{F}^2>D{G}^2$. Thus, $D{G}^2-A{F}^2$ is negative.
• Therefore, $C{F}^2-C{G}^2$ is negative, which means $C{F}^2<C{G}^2$.
• Taking the square root, CF < CG.
• This shows that the distance of the longer chord AB from the centre is less than the distance of the shorter chord DE from the centre.

Example 3:

Given: In a circle of radius 10 cm, chord AB is of length 16 cm and chord CD is of length 12 cm. To find: Which chord is farther from the centre and by what distance?

Solution:

• Let O be the centre of the circle. Radius r = 10 cm.
• For chord AB: Length AB = 16 cm. Let OE ⊥ AB.
• $AE=\frac{1}{2}\times AB=\frac{1}{2}\times 16=8$ cm.
• In right-angled ∆OEA: $O{A}^2=O{E}^2+A{E}^2$
• ${10}^2=O{E}^2+8^2$
• $100=O{E}^2+64$
• $O{E}^2=100-64=36$
• $OE=\sqrt{36}=6$ cm. (Distance of AB from centre)
• For chord CD: Length CD = 12 cm. Let OF ⊥ CD.
• $CF=\frac{1}{2}\times CD=\frac{1}{2}\times 12=6$ cm.
• In right-angled ∆OFC: $O{C}^2=O{F}^2+F{C}^2$
• ${10}^2=O{F}^2+6^2$
• $100=O{F}^2+36$
• $O{F}^2=100-36=64$
• $OF=\sqrt{64}=8$ cm. (Distance of CD from centre)
• Comparing distances: OE = 6 cm, OF = 8 cm.
• Since 6 cm < 8 cm, chord AB is closer to the centre.
• Chord CD is farther from the centre.
• The difference in distance is OF - OE = 8 cm - 6 cm = 2 cm.

Answer: Chord CD is farther from the centre by 2 cm.

Example 4:

Given: Chord AB is at a distance of 5 cm from the centre of a circle. Chord CD is at a distance of 12 cm from the centre of the same circle. To find: Which chord is longer and why?

Solution:

• We are given the distances of two chords from the centre.
• Distance of chord AB = 5 cm.
• Distance of chord CD = 12 cm.
• According to Theorem 8, the longer chord is closer to the centre.
• Since 5 cm < 12 cm, chord AB is closer to the centre than chord CD.
• Therefore, chord AB is longer than chord CD.

Answer: Chord AB is longer because it is closer to the centre.

Remember

The relationship between chord length and its distance from the centre is inversely proportional. Longer chords are closer to the centre, and shorter chords are farther away.

Exam-smart

Questions might present you with two chords and their distances from the centre, or the radius and one chord's length/distance, asking you to find the other's length or distance. You will need to apply Pythagoras theorem in conjunction with the theorems relating chord lengths and distances from the centre. Some problems might involve comparing two unequal chords and determining which is nearer or farther.

Try this: In a circle of radius 13 cm, chord AB is at a distance of 5 cm from the centre. What is the length of chord AB? Answer: 24 cm.

Angles Subtended by an Arc

An arc of a circle is a continuous part of its circumference. It's defined by two endpoints on the circle. A circle has a minor arc (the shorter path between two points) and a major arc (the longer path).

The angle subtended by an arc at the centre of the circle is the angle formed by joining the endpoints of the arc to the centre. The reflex angle is used when the arc is a major arc.

Theorem 9: Angle at Centre vs. Angle at Circumference

The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle (the circumference).

This means if an arc AB subtends an angle $\mathrm{\angle }AOB$ at the centre and $\mathrm{\angle }APB$ at a point P on the circumference, then $\mathrm{\angle }AOB=2\mathrm{\angle }APB$.

Given: Arc AB, centre O, point P on the circumference. To find: The relationship between $\mathrm{\angle }AOB$ and $\mathrm{\angle }APB$.

Example 1

Given: In a circle with centre O, arc AB subtends an angle of ${120}^{\circ }$ at the centre. To find: The angle subtended by the arc AB at any point P on the remaining part of the circle. Solution:

1. Identify the angle subtended at the centre: $\mathrm{\angle }AOB={120}^{\circ }$.
2. Use Theorem 9: $\mathrm{\angle }APB=\frac{1}{2}\mathrm{\angle }AOB$.
3. Substitute the value: $\mathrm{\angle }APB=\frac{1}{2}\times {120}^{\circ }$.
4. Calculate the result: $\mathrm{\angle }APB={60}^{\circ }$. Answer: The angle subtended by the arc AB at point P on the circle is ${60}^{\circ }$.

Example 2

Given: In the given figure (Fig. 5.26), ∠ACB is subtended by arc AB. ∠ADB is subtended by the same arc AB. ∠ACB = ${80}^{\circ }$. To find: The value of $x$, where $x=\mathrm{\angle }ADB$. Solution:

1. Identify the arc: Arc AB.
2. Identify the angles subtended by arc AB: $\mathrm{\angle }ACB$ at the centre and $\mathrm{\angle }ADB$ at the circumference.
3. Apply Theorem 9: $\mathrm{\angle }ADB=\frac{1}{2}\mathrm{\angle }ACB$.
4. Substitute the given value: $x=\frac{1}{2}\times {80}^{\circ }$.
5. Calculate $x$: $x={40}^{\circ }$. Answer: $x={40}^{\circ }$.

Corollary: Angle in a Semicircle

The angle subtended by a diameter at any point on the circumference is 90°.

Reasoning: A diameter is an arc that measures ${180}^{\circ }$ at the centre (a straight line). Applying Theorem 9, the angle at the circumference is half of this: $\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$.

Example 3

Given: Diameter AB of a circle, and point C on the circumference. To find: The measure of $\mathrm{\angle }ACB$. Solution:

1. Identify AB as a diameter. A diameter forms an arc of ${180}^{\circ }$ at the centre.
2. The angle subtended by the diameter at point C on the circumference is $\mathrm{\angle }ACB$.
3. Using the corollary of Theorem 9, the angle in a semicircle is ${90}^{\circ }$. Answer: $\mathrm{\angle }ACB={90}^{\circ }$.

Try this:

Given: An arc subtends an angle of ${150}^{\circ }$ at the centre of a circle. To find: The angle subtended by the same arc at a point on the circumference. Answer: ${75}^{\circ }$

Remember

The key is that the arc must be the same for both the centre angle and the circumference angle. Also, the point on the circumference must be on the remaining part of the circle, not on the arc itself.

Exam-smart

Expect questions where you are given the angle at the centre and asked for the angle at the circumference, or vice versa. Word problems might describe a scenario (like a Ferris wheel or a clock hand) where these angles are implicitly present. You might also be asked to prove that an angle is 90° because it's subtended by a diameter.

Concyclicity of Points

Concyclic points are points that lie on the same circle.

Theorem 10: Angles Subtended by a Line Segment

If a line segment AB subtends equal angles at two other points C and D that lie on the same side of AB, then the four points A, B, C, and D are concyclic.

Reasoning: If the angles were different, and D was not on the circle passing through A, B, C, then D would have to be either inside or outside that circle. This would lead to a contradiction where the angle subtended by AB at D would be greater than or less than the angle subtended at C, which is not the case. Therefore, D must lie on the same circle as A, B, and C.

Theorem 11: Angles of a Cyclic Quadrilateral

The sum of two opposite angles of a cyclic quadrilateral is 180°.

If ABCD is a cyclic quadrilateral, then:

Reasoning: An arc (e.g., arc BCD) subtends an angle at the circumference (∠BAD) and at the centre. The angle at the circumference is half the angle at the centre. The sum of the angles subtended by the two arcs that make up the whole circle at the centre is ${360}^{\circ }$. Therefore, the sum of opposite angles in a cyclic quadrilateral is $\frac{1}{2}\times {360}^{\circ }={180}^{\circ }$.

Example 4

Given: ABCD is a cyclic quadrilateral. $\mathrm{\angle }A={75}^{\circ }$. To find: The measure of $\mathrm{\angle }C$. Solution:

1. Identify ABCD as a cyclic quadrilateral.
2. Identify opposite angles: $\mathrm{\angle }A$ and $\mathrm{\angle }C$ are opposite angles.
3. Apply Theorem 11: $\mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }$.
4. Substitute the given value: ${75}^{\circ }+\mathrm{\angle }C={180}^{\circ }$.
5. Calculate $\mathrm{\angle }C$: $\mathrm{\angle }C={180}^{\circ }-{75}^{\circ }={105}^{\circ }$. Answer: $\mathrm{\angle }C={105}^{\circ }$.

Example 5

Given: PQRS is a cyclic quadrilateral. $\mathrm{\angle }P=(2x+10{)}^{\circ }$ and $\mathrm{\angle }R=(3x-20{)}^{\circ }$. To find: The value of $x$, and the measures of $\mathrm{\angle }P$ and $\mathrm{\angle }R$. Solution:

1. Identify PQRS as a cyclic quadrilateral.
2. Identify opposite angles: $\mathrm{\angle }P$ and $\mathrm{\angle }R$ are opposite.
3. Apply Theorem 11: $\mathrm{\angle }P+\mathrm{\angle }R={180}^{\circ }$.
4. Substitute the expressions: $(2x+10{)}^{\circ }+(3x-20{)}^{\circ }={180}^{\circ }$.
5. Simplify and solve for $x$:

$x=\frac{190}{5}=38$. 6. Calculate $\mathrm{\angle }P$: $\mathrm{\angle }P=(2\times 38+10{)}^{\circ }=(76+10{)}^{\circ }={86}^{\circ }$. 7. Calculate $\mathrm{\angle }R$: $\mathrm{\angle }R=(3\times 38-20{)}^{\circ }=(114-20{)}^{\circ }={94}^{\circ }$. 8. Verify: $\mathrm{\angle }P+\mathrm{\angle }R={86}^{\circ }+{94}^{\circ }={180}^{\circ }$. Answer: $x=38$, $\mathrm{\angle }P={86}^{\circ }$, $\mathrm{\angle }R={94}^{\circ }$.

Theorem 12: Converse of Cyclic Quadrilateral Property

If the sum of two opposite angles of a quadrilateral is 180°, then the vertices of the quadrilateral are concyclic.

Reasoning: If the vertices were not concyclic, there would be a circle passing through three vertices. The fourth vertex would either be inside or outside this circle. This would lead to a contradiction similar to the one in Theorem 10's proof, where the sum of opposite angles would not be ${180}^{\circ }$.

Try this:

Given: Quadrilateral ABCD has $\mathrm{\angle }B={100}^{\circ }$ and $\mathrm{\angle }D={90}^{\circ }$. To find: Can this quadrilateral be cyclic? Explain. Answer: No. If it were cyclic, $\mathrm{\angle }B+\mathrm{\angle }D$ should be ${180}^{\circ }$. Here, ${100}^{\circ }+{90}^{\circ }={190}^{\circ }\mathrm{\ne }{180}^{\circ }$.

Remember

The theorems about concyclicity and cyclic quadrilaterals are converses of each other. If points form a cyclic quadrilateral, their opposite angles sum to ${180}^{\circ }$. Conversely, if opposite angles of a quadrilateral sum to ${180}^{\circ }$, it is cyclic.

Exam-smart

Case studies often involve quadrilaterals inscribed in circles. You'll need to identify cyclic quadrilaterals and apply the property that opposite angles sum to ${180}^{\circ }$. Problems might give you some angles and ask for others, or provide algebraic expressions for angles to solve for unknowns, as in Example 5. Questions could also ask to prove if a given quadrilateral is cyclic based on its angle measures.