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Chapter 10 Circles (Concepts)
Prepare to delve into the elegant world of Circles, one of the most fundamental and aesthetically pleasing shapes in geometry. This chapter provides a comprehensive and rigorous study of circles and their myriad properties, moving beyond simple recognition to explore the intricate relationships between their various components like chords, arcs, and the angles they form. We will establish key theorems through logical deduction, providing you with powerful tools to analyze geometric figures involving circles and solve related problems. Understanding these properties is crucial not only in pure geometry but also in countless applications in fields like physics, engineering, design, and astronomy.
We begin by solidifying our understanding of the basic terminology associated with circles. A circle is defined as the set of all points in a plane that are equidistant from a fixed point, called the center. This fixed distance is known as the radius ($r$). A line segment passing through the center with endpoints on the circle is the diameter ($d=2r$). A chord is any line segment whose endpoints lie on the circle (the diameter is the longest possible chord). An arc is a continuous part of the circle's circumference, classified as a minor arc (shorter) or major arc (longer). A chord divides the circle's interior into two regions called segments (minor and major). The region bounded by two radii and the intercepted arc is a sector (minor and major). The total length of the boundary is the circumference ($C = 2\pi r$).
Our exploration then focuses on key theorems related to chords and their interaction with the center:
- Theorem 1: Equal chords of a circle (or of congruent circles) subtend equal angles at the center. Its converse is also true: If the angles subtended by two chords at the center are equal, then the chords are equal in length.
- Theorem 2: The perpendicular drawn from the center of a circle to a chord bisects the chord. The converse holds as well: The line drawn through the center of a circle to bisect a chord is perpendicular to the chord. This establishes a fundamental link between the center, chord, and perpendicularity.
- Theorem 3: There is one and only one circle that can pass through three given non-collinear points. (The center of this circle is the circumcenter, found at the intersection of the perpendicular bisectors of the segments joining the points).
- Theorem 4: Equal chords of a circle (or of congruent circles) are equidistant from the center. Conversely, chords that are equidistant from the center of a circle are equal in length.
Next, we investigate the crucial relationships between arcs and the angles they subtend:
- Theorem 5: The angle subtended by an arc at the center of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle. This is a cornerstone theorem linking central and inscribed angles.
- Theorem 6: Angles subtended by the same arc (or equal arcs) in the same segment of a circle are equal.
- Theorem 7: The angle inscribed in a semicircle is always a right angle ($90^\circ$). This is a direct consequence of Theorem 5, as the angle at the center subtended by the diameter (which defines the semicircle) is $180^\circ$.
- Theorem 8 (Converse related to Theorem 6): If a line segment joining two points subtends equal angles at two other points lying on the same side of the line, then these four points are concyclic (they lie on the same circle).
Finally, the chapter introduces a special type of quadrilateral intimately connected with circles: the Cyclic Quadrilateral. This is defined as a quadrilateral whose four vertices all lie on the circumference of a circle. The most significant property, which we will prove, is:
- Theorem 9: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$ (i.e., they are supplementary). The converse is also a vital tool: If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.
These nine theorems and their converses form a powerful arsenal for solving a wide variety of geometric problems involving circles, enabling us to find unknown angle measures, determine lengths, prove congruency or similarity, and establish geometric facts through rigorous logical reasoning.
Terms Related to Circles
A circle is a perfectly round two-dimensional shape. It is defined as the set (or locus) of all points in a plane that are at a fixed distance from a single fixed point. The fixed point is called the centre, and the fixed distance is the radius.
A circle divides the plane it lies on into three parts:
- The interior of the circle (the region inside the circle).
- The circle itself (the boundary).
- The exterior of the circle (the region outside the circle).
Basic Components of a Circle
1. Centre and Radius
The centre is the fixed point (usually denoted by O) that is equidistant from all points on the circle.
The radius (plural: radii) is the distance from the centre to any point on the circle. It also refers to the line segment connecting the centre to a point on the circle. All radii of a circle are equal in length.
2. Chord
A chord is a line segment whose endpoints both lie on the circle. In the figure, PQ is a chord.
3. Diameter
A diameter is a special chord that passes through the centre of the circle. It is the longest possible chord in a circle.
The length of the diameter ($d$) is always twice the length of the radius ($r$).
$d = 2r \quad \text{or} \quad r = \frac{d}{2}$
4. Arc
An arc is a continuous part of the circumference of a circle. Any two points on a circle divide it into two arcs:
- Minor Arc: The shorter arc between the two points. It is denoted by the two endpoints, like $\stackrel{\frown}{\text{PQ}}$.
- Major Arc: The longer arc between the two points. It is usually denoted by three letters to avoid ambiguity, like $\stackrel{\frown}{\text{PRQ}}$.
5. Semicircle
A semicircle is exactly half of a circle. It is the arc formed when the endpoints are the ends of a diameter. A diameter divides a circle into two equal arcs, each called a semicircle.
6. Circumference
The circumference is the total length of the boundary of the circle. It is the perimeter of the circle. The formula for the circumference (C) is:
$\text{C} = 2\pi r \quad \text{or} \quad \text{C} = \pi d$
Here, $\pi$ (pi) is a mathematical constant representing the ratio of a circle's circumference to its diameter, approximately equal to 3.14159.
Regions of a Circle
1. Segment
A segment is the region enclosed by a chord and its corresponding arc. A chord divides a circle into two segments:
- Minor Segment: The region bounded by the chord and the minor arc.
- Major Segment: The region bounded by the chord and the major arc.
2. Sector
A sector is the region enclosed by two radii and the arc between them, resembling a slice of a pie.
- Minor Sector: The region bounded by two radii and the minor arc.
- Major Sector: The region bounded by two radii and the major arc.
The angle formed by the two radii at the centre ($\angle \text{AOB}$) is called the central angle of the sector.
Lines in Relation to a Circle
1. Secant
A secant is a line that intersects a circle at two distinct points. It is essentially an extended chord.
2. Tangent
A tangent is a line that touches the circle at exactly one point. This point is called the point of contact or point of tangency.
Other Important Concepts
1. Angle Subtended by a Chord/Arc
An arc or a chord is said to subtend an angle at a point. This is the angle formed by joining the endpoints of the arc/chord to that point.
- Angle at the Centre: The angle subtended by a chord/arc at the centre of the circle (e.g., $\angle \text{POQ}$).
- Angle on the Circumference: The angle subtended by a chord/arc at any point on the remaining part of the circle (e.g., $\angle \text{PRQ}$).
2. Concentric Circles
Concentric circles are circles that share the same centre but have different radii. They are like ripples in water or an archery target.
3. Equal or Congruent Circles
Two circles are said to be congruent or equal if they have the same radius. They can be perfectly superimposed on one another.
4. Circumscribed Circle (Circumcircle)
The circumcircle of a polygon is a circle that passes through all the vertices of the polygon. The polygon is said to be inscribed in the circle. The centre of this circle is the circumcenter.
5. Inscribed Circle (Incircle)
The incircle of a polygon is a circle that is tangent to all the sides of the polygon from the inside. The polygon is said to be circumscribed about the circle. The centre of this circle is the incenter.
Congruence of Circles
In geometry, congruence means that two figures have the exact same shape and size. For circles, the shape is always the same. Therefore, the size of a circle is the only factor that determines if it is congruent to another circle.
The size of a circle is uniquely defined by its radius.
Definition of Congruent Circles
Two circles are defined as congruent if and only if their radii are equal.
Let's consider two circles, $C_1$ with centre $O_1$ and radius $r_1$, and $C_2$ with centre $O_2$ and radius $r_2$.
Circle $C_1$ is congruent to Circle $C_2$ if, and only if, their radii are equal.
$\text{Circle } C_1 \cong \text{Circle } C_2 \iff r_1 = r_2$
This means congruent circles can be perfectly placed one on top of the other, a concept known as superposition.
Implications of Congruence
If two circles are congruent, it means their radii are equal. Consequently, any measurement that depends on the radius will also be equal for both circles.
- Equal Radii: $r_1 = r_2$ (This is the definition).
- Equal Diameters: Since $d = 2r$, if $r_1 = r_2$, then $d_1 = d_2$.
- Equal Circumferences: Since $C = 2\pi r$, if $r_1 = r_2$, then $C_1 = C_2$.
- Equal Areas: Since $A = \pi r^2$, if $r_1 = r_2$, then $A_1 = A_2$.
Conversely, if any of these properties (diameter, circumference, or area) are equal for two circles, it implies their radii must be equal, and therefore the circles are congruent.
Example 1. Circle P has a diameter of 10 cm. Circle Q has a radius of 5 cm. Are the circles congruent?
Answer:
Given:
For Circle P, Diameter = 10 cm.
For Circle Q, Radius = 5 cm.
To Find:
Whether Circle P is congruent to Circle Q.
Solution:
Two circles are congruent if their radii are equal. We need to find and compare the radii of both circles.
For Circle P:
The radius is half of the diameter.
$\text{Radius of P } (r_P) = \frac{\text{Diameter}}{2} = \frac{10 \text{ cm}}{2}$
$r_P = 5 \text{ cm}$
... (i)
For Circle Q:
The radius is given directly.
$\text{Radius of Q } (r_Q) = 5 \text{ cm}$
(Given) ... (ii)
Comparing the radii from equations (i) and (ii):
$r_P = r_Q = 5 \text{ cm}$
Since the radii of Circle P and Circle Q are equal, the circles are congruent.
Therefore, Circle P is congruent to Circle Q.
Angle Subtended by a Chord at a Point
When we talk about the angle subtended by a chord at a point, we are referring to the angle formed by connecting the endpoints of the chord to that specific point. This point can be the centre of the circle or any point on the circumference of the circle.
In the figure above, AB is a chord of a circle with centre O. P is a point on the major arc AB, and Q is a point on the minor arc AB.
Angle Subtended at the Centre
The angle subtended by a chord at the centre of the circle is the angle formed by the two radii drawn from the centre to the endpoints of the chord.
For chord AB in the figure, the angle subtended by the chord at the centre O is $\angle \text{AOB}$.
Angle Subtended at a Point on the Circle (Circumference)
The angle subtended by a chord at a point on the circumference of the circle is the angle formed by joining the endpoints of the chord to that point on the circumference.
- For chord AB, if P is a point on the major arc AB, the angle subtended by the chord at P is $\angle \text{APB}$.
- For chord AB, if Q is a point on the minor arc AB, the angle subtended by the chord at Q is $\angle \text{AQB}$.
The arc AB subtends the same angle $\angle \text{AOB}$ at the centre. The arc also subtends angles at points on the remaining part of the circle (the circumference).
Theorem 10.1: Equal Chords Subtend Equal Angles at the Centre
Theorem 10.1. Equal chords of a circle subtend equal angles at the centre.
Proof:
Given:
A circle with centre O. Two chords AB and CD such that $\text{AB} = \text{CD}$.
To Prove:
The angles subtended by these chords at the centre are equal, i.e., $\angle \text{AOB} = \angle \text{COD}$.
Proof:
Consider the triangles formed by the chords and the radii to their endpoints: $\triangle \text{AOB}$ and $\triangle \text{COD}$.
In $\triangle \text{AOB}$ and $\triangle \text{COD}$, we compare their sides:
$\text{AO} = \text{CO}$
(Both are radii of the same circle)
$\text{BO} = \text{DO}$
(Both are radii of the same circle)
We are given that the chords are equal:
$\text{AB} = \text{CD}$
(Given)
Since all three sides of $\triangle \text{AOB}$ are equal to the corresponding three sides of $\triangle \text{COD}$, by the Side-Side-Side (SSS) congruence rule:
$\triangle \text{AOB} \cong \triangle \text{COD}$
(By SSS congruence)
By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding angles of congruent triangles are equal.
$\angle \text{AOB} = \angle \text{COD}$
(CPCTC)
Thus, equal chords of a circle subtend equal angles at the centre. This completes the proof.
Theorem 10.2: Converse of Theorem 10.1
Theorem 10.2. If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
Proof:
Given:
A circle with centre O. Two chords AB and CD such that the angles subtended by them at the centre are equal, i.e., $\angle \text{AOB} = \angle \text{COD}$.
To Prove:
The chords are equal in length, i.e., $\text{AB} = \text{CD}$.
Proof:
Consider the triangles formed by the chords and the radii to their endpoints: $\triangle \text{AOB}$ and $\triangle \text{COD}$.
In $\triangle \text{AOB}$ and $\triangle \text{COD}$, we compare their sides and the included angle:
$\text{AO} = \text{CO}$
(Both are radii of the same circle) ... (1)
$\text{BO} = \text{DO}$
(Both are radii of the same circle) ... (2)
We are given that the angles subtended at the centre are equal:
$\angle \text{AOB} = \angle \text{COD}$
(Given) ... (3)
We have two sides (AO and BO in $\triangle \text{AOB}$, and CO and DO in $\triangle \text{COD}$) and the angle included between them ($\angle \text{AOB}$ and $\angle \text{COD}$) which are equal.
By the Side-Angle-Side (SAS) congruence rule, if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.
$\triangle \text{AOB} \cong \triangle \text{COD}$
(By SAS congruence, using (1), (3), (2))
By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding sides of congruent triangles are equal.
$\text{AB} = \text{CD}$
(CPCTC)
Thus, if the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. This completes the proof.
Equal Chords and Their Distances from the Centre
The relationship between the length of a chord and its distance from the centre of a circle is an important concept. The distance of a chord from the centre is defined as the length of the perpendicular segment drawn from the centre to the chord. We will explore theorems related to equal chords and their distances from the centre.
In the figure above, OM represents the perpendicular distance of the chord AB from the centre O. M lies on the chord AB.
Theorem 10.3: Perpendicular from the Centre to a Chord
Theorem 10.3. The perpendicular from the centre of a circle to a chord bisects the chord.
Proof:
Given:
A circle with centre O and a chord AB. A line segment OM is drawn such that $\text{OM} \perp \text{AB}$.
To Prove:
The perpendicular OM bisects the chord AB, i.e., $\text{AM} = \text{BM}$.
Construction:
Join the centre O to the endpoints of the chord, A and B. This gives us the radii OA and OB.
Proof:
We consider the two triangles formed by the construction, $\triangle \text{OAM}$ and $\triangle \text{OBM}$.
In $\triangle \text{OAM}$ and $\triangle \text{OBM}$, we have:
$\text{OA} = \text{OB}$
(Radii of the same circle)
$\text{OM} = \text{OM}$
(Common side)
$\angle \text{OMA} = \angle \text{OMB}$
(Each is $90^\circ$, as $\text{OM} \perp \text{AB}$ is given)
The two triangles are right-angled triangles. We can apply the Right Angle-Hypotenuse-Side (RHS) congruence rule.
Here, OA and OB are the hypotenuses.
$\triangle \text{OAM} \cong \triangle \text{OBM}$
(By RHS congruence rule)
Since the triangles are congruent, their corresponding parts must be equal (CPCTC - Corresponding Parts of Congruent Triangles are equal).
Therefore,
$\text{AM} = \text{BM}$
(By CPCTC)
This shows that M is the midpoint of the chord AB.
Thus, the perpendicular from the centre of a circle to a chord bisects the chord.
Theorem 10.4: Converse of Theorem 10.3
Theorem 10.4. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Proof:
Given:
A circle with centre O. AB is a chord. A line is drawn through the centre O to the midpoint M of the chord AB. So, M is the midpoint of AB, which implies $\text{AM} = \text{BM}$. The line OM is drawn.
To Prove:
The line OM is perpendicular to the chord AB, i.e., $\text{OM} \perp \text{AB}$. This means we need to show that $\angle \text{OMA} = \angle \text{OMB} = 90^\circ$.
Construction:
Join OA and OB. OA and OB are radii of the circle.
Proof:
Consider the triangles $\triangle \text{OMA}$ and $\triangle \text{OMB}$.
We compare their sides:
$\text{OA} = \text{OB}$
(Both are radii of the same circle) ... (1)
$\text{OM} = \text{OM}$
(Common side to both triangles) ... (2)
We are given that M is the midpoint of AB:
$\text{AM} = \text{BM}$
(Given M is the midpoint of AB) ... (3)
Since all three sides of $\triangle \text{OMA}$ are equal to the corresponding three sides of $\triangle \text{OMB}$ (from equations (1), (2), and (3)), by the Side-Side-Side (SSS) congruence rule:
$\triangle \text{OMA} \cong \triangle \text{OMB}$
(By SSS congruence)
By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding angles are equal.
$\angle \text{OMA} = \angle \text{OMB}$
(CPCTC) ... (4)
The angles $\angle \text{OMA}$ and $\angle \text{OMB}$ are adjacent angles on the straight line AB and form a linear pair.
$\angle \text{OMA} + \angle \text{OMB} = 180^\circ$
(Linear Pair Axiom)
Substitute $\angle \text{OMB}$ with $\angle \text{OMA}$ from equation (4):
$\angle \text{OMA} + \angle \text{OMA} = 180^\circ$
$2\angle \text{OMA} = 180^\circ$
$\angle \text{OMA} = \frac{180^\circ}{2}$
$\angle \text{OMA} = 90^\circ$
Since $\angle \text{OMA} = 90^\circ$, it follows that $\angle \text{OMB} = 90^\circ$ as well (from equation (4)).
Thus, the line segment OM is perpendicular to the chord AB ($\text{OM} \perp \text{AB}$).
The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. This completes the proof.
Theorem 10.5: Equal Chords are Equidistant from the Centre
Theorem 10.5. Equal chords of a circle are equidistant from the centre.
Proof:
Given:
A circle with centre O. Two equal chords AB and CD such that $\text{AB} = \text{CD}$. Let OM be the perpendicular distance of chord AB from O (so $\text{OM} \perp \text{AB}$, M is on AB), and ON be the perpendicular distance of chord CD from O (so $\text{ON} \perp \text{CD}$, N is on CD).
To Prove:
The chords AB and CD are equidistant from the centre O, i.e., $\text{OM} = \text{ON}$.
Construction:
Join OA and OC. OA and OC are radii of the circle.
Proof:
Since $\text{OM} \perp \text{AB}$, by Theorem 10.3, the perpendicular from the centre to a chord bisects the chord. Thus, M is the midpoint of AB.
$\text{AM} = \frac{1}{2} \text{AB}$
(By Theorem 10.3) ... (1)
Similarly, since $\text{ON} \perp \text{CD}$, by Theorem 10.3, N is the midpoint of CD.
$\text{CN} = \frac{1}{2} \text{CD}$
(By Theorem 10.3) ... (2)
We are given that the chords are equal in length:
$\text{AB} = \text{CD}$
(Given) ... (3)
Dividing both sides of equation (3) by 2, we get $\frac{1}{2}\text{AB} = \frac{1}{2}\text{CD}$.
From equations (1) and (2), substituting $\frac{1}{2}\text{AB}$ with AM and $\frac{1}{2}\text{CD}$ with CN:
$\text{AM} = \text{CN}$
... (4)
Now, consider the right-angled triangles $\triangle \text{OMA}$ and $\triangle \text{ONC}$ (since $\angle \text{OMA} = 90^\circ$ and $\angle \text{ONC} = 90^\circ$).
In right-angled $\triangle \text{OMA}$ and $\triangle \text{ONC}$, we compare their sides:
$\text{OA} = \text{OC}$
(Both are radii of the same circle - Hypotenuses) ... (5)
$\text{AM} = \text{CN}$
(Proven in equation (4) - Sides)
And the right angles:
$\angle \text{OMA} = \angle \text{ONC} = 90^\circ$
(Given OM $\perp$ AB and ON $\perp$ CD - Right Angles)
By the Right Angle-Hypotenuse-Side (RHS) congruence rule:
$\triangle \text{OMA} \cong \triangle \text{ONC}$
(By RHS congruence)
By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding sides are equal.
$\text{OM} = \text{ON}$
(CPCTC)
Since OM and ON are the distances of chords AB and CD from the centre O, this proves that equal chords are equidistant from the centre.
Thus, equal chords of a circle are equidistant from the centre. This completes the proof.
Theorem 10.6: Converse of Theorem 10.5
Theorem 10.6. Chords equidistant from the centre of a circle are equal in length.
Proof:
Given:
A circle with centre O. Two chords AB and CD. OM is the perpendicular distance of chord AB from O ($\text{OM} \perp \text{AB}$, M is on AB), and ON is the perpendicular distance of chord CD from O ($\text{ON} \perp \text{CD}$, N is on CD). The distances are equal, i.e., $\text{OM} = \text{ON}$.
To Prove:
The chords are equal in length, i.e., $\text{AB} = \text{CD}$.
Construction:
Join OA and OC. OA and OC are radii of the circle.
Proof:
Consider the right-angled triangles $\triangle \text{OMA}$ and $\triangle \text{ONC}$ (since $\angle \text{OMA} = 90^\circ$ and $\angle \text{ONC} = 90^\circ$).
In right-angled $\triangle \text{OMA}$ and $\triangle \text{ONC}$, we compare their sides:
$\text{OA} = \text{OC}$
(Both are radii of the same circle - Hypotenuses) ... (1)
$\text{OM} = \text{ON}$
(Given the chords are equidistant) ... (2)
$\angle \text{OMA} = \angle \text{ONC} = 90^\circ$
(Given OM $\perp$ AB and ON $\perp$ CD - Right Angles)
By the Right Angle-Hypotenuse-Side (RHS) congruence rule:
$\triangle \text{OMA} \cong \triangle \text{ONC}$
(By RHS congruence, using (1), (2), and the right angles)
By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding sides are equal.
$\text{AM} = \text{CN}$
(CPCTC) ... (3)
Since $\text{OM} \perp \text{AB}$, by Theorem 10.3, the perpendicular from the centre to a chord bisects the chord. Thus, M is the midpoint of AB.
$\text{AB} = 2 \times \text{AM}$
(By Theorem 10.3)
Similarly, since $\text{ON} \perp \text{CD}$, by Theorem 10.3, N is the midpoint of CD.
$\text{CD} = 2 \times \text{CN}$
(By Theorem 10.3)
From equation (3), $\text{AM} = \text{CN}$. Multiply both sides by 2:
$2 \times \text{AM} = 2 \times \text{CN}$
Substitute AB for $2 \times$ AM and CD for $2 \times$ CN:
$\text{AB} = \text{CD}$
Thus, chords equidistant from the centre of a circle are equal in length. This completes the proof.
Example 1. In a circle with radius 5 cm, AB is a chord of length 8 cm. Find the distance of the chord from the centre.
Answer:
Given:
A circle with centre O and radius = 5 cm. Chord AB with length = 8 cm.
To Find:
The distance of the chord AB from the centre O.
Solution:
Let the centre of the circle be O. The radius is OA = 5 cm.
Let AB be the chord of length 8 cm.
To find the distance of the chord from the centre, we draw a perpendicular from the centre O to the chord AB. Let M be the foot of the perpendicular from O to AB. So, OM $\perp$ AB.
By Theorem 10.3, the perpendicular from the centre of a circle to a chord bisects the chord. This means M is the midpoint of AB.
$\text{AM} = \frac{1}{2} \times \text{AB}$
(By Theorem 10.3)
Substitute the given length of AB:
$\text{AM} = \frac{1}{2} \times 8 \text{ cm}$
$\text{AM} = 4 \text{ cm}$
... (1)
Now, consider the right-angled triangle $\triangle \text{OMA}$. We know OA (hypotenuse) and AM (one side). We want to find OM (the other side, which is the distance). We can use the Pythagoras Theorem in $\triangle \text{OMA}$.
$\text{OA}^2 = \text{OM}^2 + \text{AM}^2$
(By Pythagoras Theorem in $\triangle \text{OMA}$)
Substitute the known values OA = 5 cm and AM = 4 cm (from equation (1)):
$5^2 = \text{OM}^2 + 4^2$
$25 = \text{OM}^2 + 16$
Subtract 16 from both sides to isolate $\text{OM}^2$:
$\text{OM}^2 = 25 - 16$
$\text{OM}^2 = 9$
Take the square root of both sides. Since OM is a distance, it must be positive.
$\text{OM} = \sqrt{9}$
$\text{OM} = 3$
The distance is in centimeters.
$\text{OM} = 3 \text{ cm}$
The distance of the chord AB from the centre O is 3 cm.
Angles Subtended by Arcs and Chords at Points on the Circle
Both arcs and chords of a circle can subtend angles at different points. The angle subtended by an arc (or the chord corresponding to the arc) at the centre is the angle formed by the radii to the endpoints of the arc. The angle subtended at any point on the remaining part of the circle is formed by connecting the endpoints of the arc/chord to that point on the circumference.
In the figure, arc PQ subtends $\angle \text{POQ}$ at the centre O and $\angle \text{PRQ}$ at point R on the remaining part of the circle (major arc PRQ).
For a minor arc, the angle subtended at the centre is less than $180^\circ$. For a major arc, the angle subtended at the centre is the reflex angle formed by the radii (greater than $180^\circ$).
Theorem 10.7: Angle at the Centre and Angle at the Circumference
Theorem 10.7. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Proof:
Given:
A circle with centre O. An arc PQ subtends $\angle \text{POQ}$ at the centre and $\angle \text{PRQ}$ at a point R on the remaining part of the circle.
We need to consider three cases:
Case 1: PQ is a minor arc.
Case 2: PQ is a semicircle.
Case 3: PQ is a major arc.
To Prove:
$\angle \text{POQ} = 2 \angle \text{PRQ}$. (For Case 3, it is the reflex $\angle \text{POQ}$).
Construction:
Join R to the centre O and extend the line segment to a point S outside the circle.
Proof:
We use the property that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Consider $\triangle \text{POR}$.
$\text{OP} = \text{OR}$
(Radii of the same circle)
Since the sides are equal, the angles opposite to them are also equal.
$\angle \text{OPR} = \angle \text{ORP}$
(Angles opposite equal sides)
The angle $\angle \text{POS}$ is an exterior angle to $\triangle \text{POR}$.
$\angle \text{POS} = \angle \text{OPR} + \angle \text{ORP} = \angle \text{ORP} + \angle \text{ORP}$
$\angle \text{POS} = 2 \angle \text{ORP}$
... (i)
Similarly, consider $\triangle \text{QOR}$.
$\text{OQ} = \text{OR}$
(Radii of the same circle)
$\angle \text{OQR} = \angle \text{ORQ}$
(Angles opposite equal sides)
The angle $\angle \text{QOS}$ is an exterior angle to $\triangle \text{QOR}$.
$\angle \text{QOS} = \angle \text{OQR} + \angle \text{ORQ} = \angle \text{ORQ} + \angle \text{ORQ}$
$\angle \text{QOS} = 2 \angle \text{ORQ}$
... (ii)
Now, we apply this to the three cases.
For Case 1 (Minor Arc):
From the figure, $\angle \text{POQ} = \angle \text{POS} + \angle \text{QOS}$.
Adding equations (i) and (ii):
$\angle \text{POS} + \angle \text{QOS} = 2 \angle \text{ORP} + 2 \angle \text{ORQ}$
$\angle \text{POQ} = 2 (\angle \text{ORP} + \angle \text{ORQ})$
$\angle \text{POQ} = 2 \angle \text{PRQ}$
For Case 2 (Semicircle):
The points P, O, Q lie on a straight line. So, $\angle \text{POQ} = 180^\circ$ (a straight angle).
Using the same logic as Case 1, we get $\angle \text{POQ} = 2 \angle \text{PRQ}$.
$180^\circ = 2 \angle \text{PRQ}$
$\angle \text{PRQ} = 90^\circ$. This leads to the important corollary: the angle in a semicircle is a right angle.
For Case 3 (Major Arc):
The angle subtended by the major arc PQ at the centre is the reflex angle $\angle \text{POQ}$.
Reflex $\angle \text{POQ} = \angle \text{POS} + \angle \text{QOS}$.
From equations (i) and (ii):
Reflex $\angle \text{POQ} = 2 \angle \text{ORP} + 2 \angle \text{ORQ} = 2(\angle \text{ORP} + \angle \text{ORQ})$
Reflex $\angle \text{POQ} = 2 \angle \text{PRQ}$
In all three cases, the theorem holds true.
Thus, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Theorem 10.8: Angle in a Semicircle
Theorem 10.8. Angle in a semicircle is a right angle.
Proof:
Given:
A circle with centre O. AB is a diameter of the circle. C is any point on the semicircle formed by the diameter AB.
To Prove:
The angle subtended by the diameter at point C on the semicircle is a right angle, i.e., $\angle \text{ACB} = 90^\circ$.
Proof:
The diameter AB is a line segment passing through the centre O. The arc ACB is a semicircle.
The angle subtended by the diameter AB at the centre O is the angle $\angle \text{AOB}$. Since AB is a straight line segment,
$\angle \text{AOB} = 180^\circ$
(Angle on a straight line)
The arc AB subtends the angle $\angle \text{AOB}$ at the centre O and the angle $\angle \text{ACB}$ at the point C on the remaining part of the circle (which is the semicircle itself in this context, considered as the arc from A to B in one direction).
By Theorem 10.7, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\angle \text{AOB} = 2 \angle \text{ACB}$
(By Theorem 10.7)
Substitute the value of $\angle \text{AOB}$:
$180^\circ = 2 \angle \text{ACB}$
Divide both sides by 2:
$\angle \text{ACB} = \frac{180^\circ}{2}$
$\angle \text{ACB} = 90^\circ$
Thus, the angle in a semicircle is a right angle. This completes the proof.
Theorem 10.9: Angles in the Same Segment
Theorem 10.9. Angles in the same segment of a circle are equal.
Proof:
Given:
A circle with centre O. Chord AB. Points C and D are any two points on the same segment of the circle formed by chord AB (i.e., C and D are both on the major arc AB or both on the minor arc AB).
To Prove:
The angles subtended by the chord AB at points C and D in the same segment are equal, i.e., $\angle \text{ACB} = \angle \text{ADB}$.
Construction:
Join OA and OB to form the angle subtended by the arc AB at the centre.
Proof:
The arc AB subtends the angle $\angle \text{AOB}$ at the centre O and the angle $\angle \text{ACB}$ at point C on the remaining part of the circle.
By Theorem 10.7, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\angle \text{AOB} = 2 \angle \text{ACB}$
(By Theorem 10.7) ... (1)
Similarly, the same arc AB subtends the angle $\angle \text{AOB}$ at the centre O and the angle $\angle \text{ADB}$ at point D on the remaining part of the circle.
By Theorem 10.7:
$\angle \text{AOB} = 2 \angle \text{ADB}$
(By Theorem 10.7) ... (2)
From equation (1) and equation (2), since both $2 \angle \text{ACB}$ and $2 \angle \text{ADB}$ are equal to $\angle \text{AOB}$:
$2 \angle \text{ACB} = 2 \angle \text{ADB}$
(Things which are equal to the same thing are equal to one another)
Divide both sides by 2:
$\angle \text{ACB} = \angle \text{ADB}$
Thus, angles in the same segment of a circle are equal. This completes the proof. This applies whether the segment is major or minor.
Theorem 10.10: Condition for Four Points to be Concyclic
Theorem 10.10. If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, then the four points lie on a circle (i.e., they are concyclic).
Proof:
Given:
Four points A, B, C, and D, such that C and D are on the same side of the line segment AB.
The angles subtended by the segment AB at C and D are equal, i.e., $\angle \text{ACB} = \angle \text{ADB}$.
To Prove:
The four points A, B, C, and D are concyclic (lie on the same circle).
Proof (by Contradiction):
We know that a unique circle can be drawn through any three non-collinear points. Let's draw a circle that passes through the points A, B, and C.
Now, let's assume that the fourth point, D, does not lie on this circle. If D is not on the circle, it must either be in the interior or the exterior of the circle.
Case 1: D lies in the exterior of the circle.
Produce the line segment AD to intersect the circle at a point D'. Join B to D'.
Since points C and D' lie on the same circle, the angles subtended by the same arc AB in the same segment must be equal (Theorem 10.9).
$\angle \text{ACB} = \angle \text{AD'B}$
(Angles in the same segment)
But we are given that:
$\angle \text{ACB} = \angle \text{ADB}$
(Given)
Therefore, it must be that $\angle \text{AD'B} = \angle \text{ADB}$.
However, in $\triangle \text{BDD'}$, the exterior angle $\angle \text{AD'B}$ must be greater than the interior opposite angle $\angle \text{BDD'}$ (which is the same as $\angle \text{BDA}$ or $\angle \text{ADB}$).
So, $\angle \text{AD'B} > \angle \text{ADB}$. This contradicts our finding that $\angle \text{AD'B} = \angle \text{ADB}$.
Thus, our assumption that D lies in the exterior of the circle is false.
Case 2: D lies in the interior of the circle.
Let the line AD intersect the circle at D' when extended. Join B to D'.
Again, $\angle \text{ACB} = \angle \text{AD'B}$ (Angles in the same segment) and $\angle \text{ACB} = \angle \text{ADB}$ (Given). This implies $\angle \text{AD'B} = \angle \text{ADB}$.
But in $\triangle \text{BD'D}$, the exterior angle $\angle \text{ADB}$ must be greater than the interior opposite angle $\angle \text{AD'B}$.
So, $\angle \text{ADB} > \angle \text{AD'B}$. This again contradicts our finding that the angles are equal.
Thus, our assumption that D lies in the interior of the circle is also false.
Since D cannot lie inside or outside the circle, it must lie on the circle.
Therefore, the four points A, B, C, and D are concyclic.
Cyclic Quadrilateral
A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a single circle. These quadrilaterals have a special property related to their angles.
In the figure, ABCD is a cyclic quadrilateral as all its vertices A, B, C, and D lie on the circumference of the circle.
Theorem 10.11: Opposite Angles of a Cyclic Quadrilateral
Theorem 10.11. The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.
Proof:
Given:
ABCD is a cyclic quadrilateral inscribed in a circle with centre O.
To Prove:
(i) $\angle \text{DAB} + \angle \text{BCD} = 180^\circ$
(ii) $\angle \text{ABC} + \angle \text{ADC} = 180^\circ$
Construction:
Join the centre O to the vertices B and D.
Proof:
We will use Theorem 10.7, which states that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Consider the minor arc DCB. It subtends $\angle \text{DOB}$ at the centre and $\angle \text{DAB}$ at point A on the remaining part of the circle.
$\angle \text{DOB} = 2 \angle \text{DAB}$
... (i)
Now, consider the major arc DAB. It subtends the reflex angle $\angle \text{DOB}$ at the centre and $\angle \text{DCB}$ at point C on the remaining part of the circle.
Reflex $\angle \text{DOB} = 2 \angle \text{DCB}$
... (ii)
The sum of an angle and its reflex angle around a point is $360^\circ$.
$\angle \text{DOB} + \text{Reflex } \angle \text{DOB} = 360^\circ$
Substituting the values from equations (i) and (ii):
$2 \angle \text{DAB} + 2 \angle \text{DCB} = 360^\circ$
Dividing by 2, we get:
$\angle \text{DAB} + \angle \text{DCB} = 180^\circ$
This proves the first part.
For the second part, we use the angle sum property of a quadrilateral: the sum of all four interior angles is $360^\circ$.
$\angle \text{DAB} + \angle \text{ABC} + \angle \text{BCD} + \angle \text{ADC} = 360^\circ$
$(\angle \text{DAB} + \angle \text{BCD}) + (\angle \text{ABC} + \angle \text{ADC}) = 360^\circ$
We just proved that $(\angle \text{DAB} + \angle \text{BCD}) = 180^\circ$. Substituting this:
$180^\circ + (\angle \text{ABC} + \angle \text{ADC}) = 360^\circ$
$\angle \text{ABC} + \angle \text{ADC} = 360^\circ - 180^\circ$
$\angle \text{ABC} + \angle \text{ADC} = 180^\circ$
This proves the second part. Thus, opposite angles of a cyclic quadrilateral are supplementary.
Theorem 10.12: Converse of Theorem 10.11
Theorem 10.12. If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.
Proof:
Given:
A quadrilateral ABCD where the sum of a pair of opposite angles is $180^\circ$. Let's assume $\angle \text{B} + \angle \text{D} = 180^\circ$.
To Prove:
The quadrilateral ABCD is cyclic.
Proof (by Contradiction):
Let's draw a circle passing through three non-collinear points A, B, and C.
Assume that the fourth point, D, does not lie on this circle. If so, the line CD (or its extension) will intersect the circle at a different point, say D'.
Since A, B, C, and D' lie on the circle, ABCD' is a cyclic quadrilateral.
From Theorem 10.11, the opposite angles of a cyclic quadrilateral are supplementary.
$\angle \text{B} + \angle \text{AD'C} = 180^\circ$
... (i)
But we are given that:
$\angle \text{B} + \angle \text{ADC} = 180^\circ$
(Given) ... (ii)
Comparing equations (i) and (ii), we get:
$\angle \text{AD'C} = \angle \text{ADC}$
Now, consider $\triangle \text{AD'D}$. The exterior angle at D' is $\angle \text{AD'C}$. It must be equal to the sum of the interior opposite angles, $\angle \text{DAD'}$ and $\angle \text{ADD'}$.
$\angle \text{AD'C} = \angle \text{DAD'} + \angle \text{ADD'}$
This implies $\angle \text{AD'C} > \angle \text{ADD'}$, which is the same as $\angle \text{AD'C} > \angle \text{ADC}$.
This contradicts our finding that $\angle \text{AD'C} = \angle \text{ADC}$. The contradiction can only be resolved if D and D' are the same point.
Therefore, our initial assumption was wrong. Point D must lie on the circle.
Hence, the quadrilateral ABCD is cyclic.
Example 1. In a cyclic quadrilateral ABCD, if $\angle \text{A} = 70^\circ$, find the measure of $\angle \text{C}$.
Answer:
Given:
ABCD is a cyclic quadrilateral with $\angle \text{A} = 70^\circ$.
To Find:
The measure of $\angle \text{C}$.
Solution:
In a cyclic quadrilateral, opposite angles are supplementary (add up to $180^\circ$).
The angle opposite to $\angle \text{A}$ is $\angle \text{C}$.
$\angle \text{A} + \angle \text{C} = 180^\circ$
Substituting the given value of $\angle \text{A}$:
$70^\circ + \angle \text{C} = 180^\circ$
Solving for $\angle \text{C}$:
$\angle \text{C} = 180^\circ - 70^\circ$
$\angle \text{C} = 110^\circ$
The measure of angle C is $110^\circ$.