A closed figure obtained by joining all those points in a plane at a constant distance from a fixed point in the same plane is known as a circle. In other words, a circle is the locus of a point that moves in a plane so that its distance from a fixed point in the same plane always remains constant. The fixed point is called the centre, and the constant distance is called the circle’s radius. An angle formed between the radii, chords, or tangents of a circle is known as the angle of a circle.

In this article, we will study some properties related to the angle of a circle and theorems based on the angle properties.

Circles

As mentioned above, the set of all points in a plane at a fixed distance from a fixed point is known as a circle. The fixed point is called the centre, the fixed distance is called the circle’s radius, and its perimeter is called its circumference.

Concentric Circles

Two or more circles are said to be concentric if they have the same centre and different radii. In the below figure, \(O\) is the centre of each circle drawn, so the circles are concentric.

Equal Circles

Circles are said to be equal or congruent if they have equal radii.

Circumscribed Circle

The circumscribed circle is the circle that passes through all the vertices of a polygon. The centre of the circumscribed circle is called the circumcentre, and the polygon is called the inscribed polygon, as shown below.

Inscribed Circle

The inscribed circle touches all the sides of a polygon. It is also known as an in-circle of the polygon. The centre of the inscribed circle is called incentre, and the polygon is called a circumscribed polygon.

Chord

A chord is the line segment joining any two points on the circumference of the circle. A chord that passes through the centre of the circle is called the diameter, and it is the longest chord of a circle.

A straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is at right angles to the chord.

In the below figure, line \(OM\) drawn from the centre \(O\) to bisect the chord \(AB\) is perpendicular to \(AB.\)
i.e. \(AM = BM \Rightarrow OM \bot AB \Rightarrow \angle OMA = \angle OMB = {90^ \circ }\)
The perpendicular to a chord, from the centre of the circle, bisects the chord.
In the below figure, \(O\) is the centre of the circle, and \(OP\) is perpendicular to the chord \(AB.\)
\( \Rightarrow AP = BP\)
Greater is the size of a chord; lesser is its distance from the centre and vice-versa.
The below figure shows a circle with centre \(O.OP\) is perpendicular to chord \(AB,\) and \(OQ\) is perpendicular to chord \(CD.\) Since chord \(AB\) is greater than chord \(CD\)
\( \Rightarrow AB\) is at a smaller distance from the centre as compared to \(CD,\) i.e., \(OP < OQ.\) Conversely, as \(OP < OQ \Rightarrow AB < CD.\)
Equal chords of a circle are equidistant from the centre. Chords of a circle that are equidistant from the centre of the circle are equal.
In the below figure, if \(AB = CD \Rightarrow OM = ON\) and if \(OM = ON \Rightarrow AB = CD\)

Arcs

An arc is a part of the circumference of a circle. A chord divides the circumference of a circle into two parts, and each part is called an arc.

In the below figure, chord \(AB\) divides the circumference into two unequal arcs \(APB\) and \(AQB.\)
The arc \(APB,\) less than the semi-circle, is called the minor arc, and the arc \(AQB,\) greater than the semi-circle, is called the major arc.

Segment

A segment is the part of a circle bounded by an arc and a chord.

Angle Properties of Circle

Theorem 1: The angle which an arc of a circle subtends at the centre is double that it subtends at any point on the remaining part of the circumference.
Given: Consider a circle with centre \(O.\) Arc \(APB\) subtends angle \(AOB\) at the centre and angle \(ACB\) at point \(C\) on the remaining circumference.
To prove: \(\angle AOB = 2\angle ACB\)
Construction: Join \(CO\) and produce it to a point \(D.\)

Proof: In \(\Delta AOC\)
\(OA = OC\) (Radii of the same circle)
\(\angle OAC = \angle OCA\) (Angles opposite to equal sides of a triangle are equal) \(…..\left({\text{i}} \right)\)
\(\angle AOD = \angle OAC + \angle OCA\) (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
\( \Rightarrow \angle AOD = \angle OCA + \angle OCA\) (From \(\left({\text{i}} \right)\))
\( = 2\angle OCA…..\left({{\text{ii}}} \right)\)
Similarly, in \(\Delta BOC\)
Exterior \(\angle BOD = 2\angle OCB……..\left({{\text{iii}}} \right)\)
Now, \(\angle AOB = \angle AOD + \angle BOD\)
\( = 2\angle OCA + 2\angle OCB\) (From \(\left({{\text{ii}}} \right)\) and \(\left({{\text{iii}}} \right)\))
\( = 2\left({\angle OCA + \angle OCB} \right)\)
\( = 2\angle ACB\)

Theorem 2: Angles in the same segment of a circle are equal.
Given: A circle with centre \(O.\angle ACB\) and \(\angle ADB\) are in the same segment.
To prove: \(\angle ACB = \angle ADB\)
Construction: Join \(OA\) and \(OB\)

Proof: Arc \(APB\) subtends angle \(AOB\) at the centre and angle \(ACB\) at point \(C\) of the remaining circumference.
Therefore, \(\angle AOB = 2\angle ACB\) (Angle at the centre is twice the angle at remaining circumference) \(……\left({\text{i}} \right)\)
Similarly, \(\angle AOB = 2\angle ADB……\left({{\text{ii}}} \right)\)
\( \Rightarrow \angle ACB = \angle ADB\) (From \(\left({\text{i}} \right)\) and \(\left({\text{i}} \right)\))

Theorem 3: The angle in a semi-circle is a right angle.
Given: A circle with centre \(O.AB\) is diameter, and \(ACB\) is the angle of the semi-circle.

To prove \(\angle ACB = {90^ \circ }\)
Proof: Arc \(APB\) subtends \(\angle AOB\) at the centre and \(\angle ACB\) at point \(C\) of the remaining circumference.
Therefore, \(\angle AOB = 2\angle ACB\) (Angle at the centre is twice the angle at remaining circumference) \(…..\left({\text{i}} \right)\)
\(\angle AOB ={180^ \circ }\) (\(AOB\) is a straight line) \(…..\left({{\text{ii}}} \right)\)
\(2\angle ACB = {180^ \circ }\)
\( \Rightarrow \angle ACB = {90^ \circ }\)

Cyclic Properties

When a quadrilateral is inscribed in a circle, i.e., when the quadrilateral’s vertices lie on the circumference of a circle, the quadrilateral is called a cyclic quadrilateral.

The points, which lie on the circumference of the same circle, are called concyclic points.

Theorem 1: The opposite angles of a cyclic quadrilateral (quadrilateral inscribed in a circle) are supplementary.
To prove: \(\angle ABC + \angle ADC = {180^ \circ }\) and \(\angle BAD + \angle BCD = {180^ \circ }\)
Construction: Join \(OA\) and \(OC.\)

Proof: Arc \(ABC\) subtends angle \(AOC\) at the centre and angle \(ADC\) at point \(D\) of the remaining circumference.
Therefore, \(\angle AOC = 2\angle ADC\) (Angle at the centre is twice the angle at remaining circumference)
\( \Rightarrow \angle ADC = \frac{1}{2}\angle AOC\)
Similarly, \(\angle ABC = \frac{1}{2}\) reflex \(\angle AOC\)
Now, \(\angle ABC + \angle ADC = \frac{1}{2}\)(reflex \(\angle AOC + \angle AOC\))
\( \Rightarrow \angle ABC + \angle ADC = \frac{1}{2} \times {360^ \circ } = {180^ \circ }\)
Similarly, \(\Rightarrow \angle BAD + \angle BCD = {180^ \circ }\)

Theorem 2: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
Given: A cyclic quadrilateral \(ABCD\) whose side \(AB\) is produced to a point \(E.\)
To prove: Ext. \(\angle CBE = \angle ADC\)

Proof:
\(\angle ABC + \angle CBE = {180^ \circ }\)
\(\angle ABC + \angle ADC ={180^ \circ }\) (Opposite angles of a cyclic quadrilateral are supplementary)
Therefore, \(\angle ABC + \angle CBE = \angle ABC + \angle ADC\)
\( \Rightarrow \angle CBE = \angle ADC\)

Solved Examples – Angle Properties of Circle

Q.1. In the below figure, \(\angle AOC = {110^ \circ }.\)Calculate \(\angle ADC,\angle ABC\) and \(\angle OAC.\)

Ans: \(\angle ADC = \frac{1}{2}\angle AOC\) (Angle at the centre is twice the angle at remaining circumference)
\( = \frac{1}{2} \times {110^ \circ } = {55^ \circ }\)
\(\angle ABC + \angle ADC = {180^ \circ }\) (Opposite angles of a cyclic quadrilateral are supplementary)
\( \Rightarrow \angle ABC + {55^ \circ } = {180^ \circ }\)
\( \Rightarrow \angle ABC = {180^ \circ } – {55^ \circ } = {125^ \circ }\)
In \(\Delta AOC,\)
\(OA = OC\) (Radii of the same circle)
\(\angle OAC = \angle OCA\) (Angles opposite to equal sides)
Also, \(\angle OAC + \angle OCA + \angle AOC = {180^ \circ }\)
\( \Rightarrow 2\angle OAC + {110^ \circ } = {180^ \circ }\)
\( \Rightarrow 2\angle OAC = {70^ \circ }\)
\( \Rightarrow \angle OAC = {35^ \circ }\)
Therefore, \(\angle ABC = {125^ \circ },\angle ADC = {55^ \circ }\) and \(\angle OAC ={35^ \circ }\)

Q.2. In the below figure, \(PQ = PR\) and \(\angle PRQ = {70^ \circ }.\) Find \(\angle QAR.\)

Ans: In \(\Delta PQR,\)
\(PQ = PR\) (Given)
\(\angle PQR = \angle PRQ = {70^ \circ }\) (Angles opposite to equal sides are equal)
Also, \(\angle PQR + \angle PRQ + \angle QPR ={180^ \circ }\)
\( \Rightarrow \angle QPR = {180^ \circ } – {70^ \circ } – {70^ \circ }\)
\( \Rightarrow \angle QPR = {40^ \circ }\)
Now, \( \Rightarrow \angle QAR = \angle QPR\) (Angles of the same segment are equal)
Therefore, \(\angle QAR = {40^ \circ }\)

Q.3. The given figure shows a circle through the points \(A,B,C\) and \(D.\) If \(\angle BAC = {67^ \circ },\) find \(\angle DBC + \angle DCB.\)

Ans: \(\angle BDC = \angle BAC = {67^ \circ }\) (Angles of the same segment are equal)
In \(\Delta DBC,\)
\(\angle BDC + \angle DBC + \angle DCB = {180^ \circ }\)
\( \Rightarrow {67^ \circ } + \angle DBC + \angle DCB = {180^ \circ }\)
\( \Rightarrow \angle DBC + \angle DCB = {113^ \circ }\)

Q.4. In the given figure, \(BC||DE\) and \(O\) is the centre of the circle. If \(\angle CDE = {x^ \circ },\) find \(\angle BAC\) in terms of \({x^ \circ }.\)

Ans: Given: \(BC||DE\) and \(DC\) is a transversal.
Therefore, \(\angle BCD + \angle CDB = {x^ \circ }\) (Alternate angles)
Now, join \(OB\) and \(OC.\)

We know that the angle at the centre is twice the angle at the remaining circumference \( \Rightarrow \angle COE = 2\angle CDE = 2{x^ \circ }\) and,
\(\angle BOD = 2\angle BCD = 2{x^ \circ }\)
Also, \(DOE\) is a straight line
\( \Rightarrow \angle DOB + \angle BOC + \angle COE = {180^ \circ }\)
\( \Rightarrow 2{x^ \circ } + \angle BOC + 2{x^ \circ } = {180^ \circ }\)
\( \Rightarrow \angle BOC = {\left({180 – 4x} \right)^ \circ }\)
Also \(\angle BOC = 2\angle BAC \Rightarrow {\left({180 – 4x} \right)^ \circ } = 2\angle BAC\)
\( \Rightarrow \angle BAC = \frac{{{{\left({180 – 4x} \right)}^ \circ }}}{2} = {\left({90 – 2x} \right)^ \circ }\)

Q.5. In the below figure, \(AC\) is the diameter of the circle. \(AB = BC\) and \(\angle AED = {118^ \circ }.\) Calculate \(\angle DEC\) and \(\angle DAB\)

Ans: Join \(EC.\)

\( \Rightarrow \angle AEC = {90^ \circ }\) (Angle of a semi-circle)
\( \Rightarrow \angle DEC + \angle AEC = {118^ \circ }\) \(\left[{\angle AED = \angle DEC + \angle AEC} \right]\)
\( \Rightarrow \angle DEC = {118^ \circ } – {90^ \circ }\)
\( \Rightarrow \angle DEC = {28^ \circ }\)
Join \(AD\)
\(\angle DAC = \angle DEC = {28^ \circ }\) (Angles of the same segment are equal)
Also, \(\angle ABC = {90^ \circ }\) (Angles of semi-circle)
\(\angle BAC = \angle BCA = {45^ \circ }\left[{AB = BC} \right]\)
So, \(\angle DAB = \angle DAC + \angle BAC = {28^ \circ } + {45^ \circ } = {73^ \circ }\)

Summary

In this article, we have studied the meaning of the angle of a circle, properties related to the angle of a circle, and cyclic properties. Also, we have solved some example problems based on the angle properties of a circle and cyclic properties.

Learn the Properties of Circle

Frequently Asked Questions (FAQs)

Q.1. What is the angle of a circle?
Ans: An angle formed between the radii, chords, or tangents of a circle is known as the angle of a circle.

Q.2. What are the rules for angles in a circle?
Ans: Below is the angle properties or rules for angles in a circle.
1. The angle at which an arc of a circle subtends at the centre is double that it subtends at any point on the remaining part of the circumference.
2. Angles in the same segment of a circle are equal.
3. The angle in a semi-circle is a right angle.

Q.3. What are the cyclic properties of a quadrilateral?
Ans: 1. The opposite angles of a cyclic quadrilateral (quadrilateral inscribed in a circle) are supplementary
2. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Q.4. What are cyclic quadrilaterals?
Ans: When a quadrilateral is inscribed in a circle, i.e., the quadrilateral’s vertices lie on the circumference of a circle. This quadrilateral is called a cyclic quadrilateral.

Q.5. What are concyclic points?
Ans: The points, which lie on the circumference of the same circle, are called concyclic points.

We hope this detailed article on the angle properties of circles helped you in your studies. If you have any doubts, queries, or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!