Properties of Circle: Definition, Formulas & Solved Examples

Properties of Circle: A circle is a two-dimensional shape created by a series of points in the plane that are at a constant or fixed distance (radius) from a fixed point (centre). The fixed point is known as the origin or centre of the circle, and the fixed distance between the points is known as the radius. A circle is a plane figure with a curved surface.

In Mathematics, there are numerous properties of circles focusing on geometry. Its relationship to straight lines, polygons, and angles may also be shown. All of these facts are characteristics of the circle when taken together. Now that we know what is a circle, let us try to study the properties of circles that will clear your understanding of the theorems of circles in class 9. We will also answer what is a tangent of a circle, a chord of the circle, and many more. Scroll down to find more.

What is Circle: Properties of Circle Related to Chord

In geometry, a circle can be defined as a closed, two-dimensional curved shape. Every point on the circle is equidistant from a certain point known as the circle’s centre. A circle is a two-dimensional form that is measured in terms of radius. The term “circle” is derived from the Greek word “kirkos,” which means “ring” or “hoop.” Following are theorems of circles class 9:

Theorem 1: A perpendicular from the centre of the circle to the chord bisects the chord.

Given, a circle with centre \(O.\,AB\) is a chord such that \(OM \bot AB\)
To prove: \(AM = BM\)
Proof: In \(\Delta AOM\) and \(\Delta BOM\)
\(AO = OB\) (Radius)
\(OM = OM\) (Common)
\(\angle AMO = \angle BMO = {90^{\rm{o}}}\)
\(\therefore \Delta AOM \cong \Delta BOM\) (By RHS congruency rule)
\(\therefore AM = BM\) (By CPCT)

Theorem 2: The line segment joining the centre of the circle and midpoint of a chord is perpendicular to the chord.

Given, a circle with centre \(O.\,AB\) is a chord such that \(OM \bot AB\)
To prove: \(AM = BM\)
Proof: In \(\Delta AOM\) and \(\Delta BOM\)
\(AO = OB\) (Radius)
\(OM = OM\) (Common)
\(AM = BM\) (Given)
\(\therefore \Delta AOM \cong \Delta BOM\) (By SSS rule)
\(\therefore \angle AMO = \angle BMO = {90^{\rm{o}}}\) (By CPCT)

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Properties of a Circle Related to Angles

Theorem 1: The equal chords of a circle subtend equal angles at the centre.

Given: a circle with centre \(O.\,AB\) and \(CD\) are equal chords of a circle, i.e., \(AB = CD.\)
To prove: \(\angle AOB = \angle COD\)
Proof: In \(\Delta AOB\) and \(\Delta COD\)
\(AO = OD\) (Radius)
\(AB = CD\) (Given)
\(OB = OC\) (Radius)
\(\therefore \Delta AOB \cong \Delta DOC\) (By SSS rule)
\(\therefore \angle AOB = \angle DOC\) (By CPCT)

Theorem 2:If the angles subtended by chords of a circle at the centre are equal, then the chords are equal.

To prove: \(AB = CD\)
Proof: In \(\Delta AOB\) and \(\Delta DOC\)
Given \(\angle AOB = \angle COD\)
\(AO = OD\) (Radius)
\(\angle AOB = \angle DOC\) (Given)
\(OB = OC\) (Radius)
\(\therefore \Delta AOB \cong \Delta DOC\) (By SAS rule)
\(\therefore AB = CD\) (By CPCT)

Theorem 3: Angle subtended by a chord at the centre is double the angle at any point on the circumference.

Given: an arc \(PQ\) of a circle subtending angles \(\angle POQ\) at the centre \(O\) and \(\angle PAQ\) and a point \(A\) on the remaining part of the circle.
To prove: \(\angle POQ = 2\angle PAQ.\)
Proof: Consider the three different cases:

Case 1: When arc \(PQ\) is minor
Case 2: When arc \(PQ\) is a semicircle and
Case 3: When arc \(PQ\) is major
Join \(AO\) and extend it to a point \(B.\)
In all the cases, \(\angle BOQ = \angle OAQ + \angle AQO\) (an exterior angle of a triangle is equal to the sum of the two interior opposite angles)

Also in \(\Delta OAQ,\)
\(OA = OQ\)( Radii of a circle)
Therefore \(\angle OAQ = \angle OQA\)
\( \Rightarrow \angle BOQ = 2\angle OAQ\) ….(i)
Similarly, \(\angle BOP = 2\angle OAP\)…..(ii)

From equation (i) and (ii),
\(\angle BOP + \angle BOQ = 2\left( {\angle OAP + \angle OAQ} \right)\)
\(\angle POQ = 2\angle PAQ\)…(iii)
For case \(3,\) where \(PQ\) is the major arc, (iii) is replaced by reflex angle \(\angle POQ = 2\angle PAQ\)

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Theorem 4: Angles subtended by a chord at different points on the same side of the circumference are the same in measure. 

Given that points, \(A\) and \(B\) on this circle subtend angles \(\angle APB\) and \(\angle AQB\) points \(P\) and \(Q\) respectively.
To prove: \(\angle APB = \angle AQB\)
The angle subtended by an arc at the centre is double the angle subtended by any point on the remaining part of the circle.
\(\angle AOB = 2\angle APB\) …(i)
\(\angle AOB = 2\angle AQB\) …(ii)
From equation (i) and (ii)
\(2\angle APB = 2\angle AQB\)
\(\angle APB = \angle AQB\)

Theorem 5: The angle subtended by the diameter at the circumference is a right angle.

Proof: \(AOB\) is a straight line passing through centre \(O.\)
\(\therefore \) Angle subtended by arc \(AB\) at \(O\) is
\(\angle AOB = {180^{\rm{o}}}\)
The angle subtended by an arc at the centre is double the angle subtended by any point on the remaining part of the circle.
\(\therefore \angle AOB = 2\angle APB\)
\(\angle APB = \frac{1}{2}\angle AOB\)
\(\angle APB = \frac{1}{2} \times {180^{\rm{o}}}\)
\(\angle APB = {90^{\rm{o}}}\)

Properties of Circle Related to Tangent

Firstly, what is a tangent of a circle? A line that touches the circle at a single point is known as a tangent to a circle. The tangent is always perpendicular to the radius of the circle. Following are the theorems of circles related to tangents.

Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle \(C\left( {0,r} \right)\) and a tangent \(l\) at point \(A.\)
To prove: \(OA \bot l\)
Construction: Take a point \(B,\) other than \(A,\) on the tangent \(l.\) Join \(OB.\) Suppose \(OB\) meets the circle at point \(C.\)
Proof: We know that, among all line segments joining the point \(O\) i.e. centre of the circle to a point on \(l\) (\(l\) is the tangent to the circle), the perpendicular is shortest to \(l.\)
\(OA = OC\) (Radius of the same circle)
Now, \(OB = OC + BC\)
\(\therefore OB > OC\)
\( \Rightarrow OB > OA\)
\( \Rightarrow OA < OB\)
\(B\) is an arbitrary point on the tangent \(l.\) Thus, \(OA\) is shorter than any other line segment joining \(O\) to any point on \(l.\)
Here, \(OA \bot l\)
This the converse of the tangent theorem.

Theorem 2: A line perpendicular to the radius at its point on the circle is a tangent to the circle.

Given: \(M\) is the centre of a circle, seg \(MN\) is a radius.
Line \(l \bot \) seg \(MN\) at \(N.\)
To prove: Line \(l\) is a tangent to the circle.
Construction: Take any point \(P,\) other than \(N,\) on the line \(l.\) Draw seg \(MP.\)
Proof: Now in \(\Delta MNP,\,\angle N\) is a right angle.
Therefore, seg \(MP\) is the hypotenuse.
Therefore, seg \(MP > \) seg \(MN.\)
As seg \(MN\) is the radius, point \(P\) cannot be on the circle.
So, no other point except point \(N,\) of line \(l\) is on the circle.
Line \(l\) intersects the circle in only one point \(N.\)
Therefore, line \(l\) is a tangent to the circle.

Theorem 3: The lengths of tangents drawn from an external point to a circle are equal.

Given: a circle with centre \(O,\) a point \(P\) lying outside the circle and two tangents \(PA,\, PB\) on the circle from \(P.\)
To prove: \(PA = PB\)
Construction: Join \(OA,\,OB\) and \(OP.\)
Proof: \(\angle OAP = \angle OBP = {90^{\rm{o}}}\) (Because the tangent at any point of a circle is perpendicular to the radius through the point of contact)
Now in right triangles, \(\Delta OAP\) and \(\Delta OBP\)
\(OA = OB\) (Radii of the same circle)
\(OP = OP\) (Common)
Therefore, \(\Delta OAP \cong \Delta OBP\) (By RHS congruency criteria)
Hence, \(PA = PB\) (By CPCT)

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Properties of Circle Related to Cyclic Quadrilateral

Theorem 1: In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary.

Given : \(ABCD\) is a cyclic quadrilateral of a circle with centre at \(O.\)
To prove : \(\angle BAD + \angle BCD = {180^{\rm{o}}}\)
\(\angle ABC + \angle ADC = {180^{\rm{o}}}\)
Proof: \(AB\) is the chord of the circle
\(\angle 5 = \angle 8\)…(i) (Angles in the same segment are equal)
\(BC\) is the chord of the circle
\(\angle 1 = \angle 6\)….(ii) (Angles in the same segment are equal)
\(CD\) is the chord of the circle
\(\angle 2 = \angle 4\)…..(iii) (Angles in the same segment are equal)
\(AD\) is the chord of the circle
\(\angle 7 = \angle 3\)….(iv) (Angles in the same segment are equal)
By angle sum property of a quadrilateral
\(\angle A + \angle B + \angle C + \angle D = {360^{\rm{o}}}\)
\(\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 7 + \angle 8 + \angle 5 + \angle 6 = {360^{\rm{o}}}\)
\(\left( {\angle 1 + \angle 2 + \angle 7 + \angle 8} \right) + \left( {\angle 3 + \angle 4 + \angle 5 + \angle 6} \right) = {360^{\rm{o}}}\)
\(\left( {\angle 1 + \angle 2 + \angle 7 + \angle 8} \right) + \left( {\angle 7 + \angle 2 + \angle 8 + \angle 1} \right) = {360^{\rm{o}}}\)

From equation (i), (ii), (iii) and (iv)
\(2\left( {\angle 1 + \angle 2 + \angle 7 + \angle 8} \right) = {360^{\rm{o}}}\)
\(\left( {\angle 1 + \angle 2 + \angle 7 + \angle 8} \right) = {180^{\rm{o}}}\)
\(\angle BAD + \angle BCD = {180^{\rm{o}}}\)
Similarly, \(\angle ABC + \angle ADC = {180^{\rm{o}}}\)
Hence proved.

Theorem 2: If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Draw a quadrilateral \(PQRS.\)
Since in quadrilateral \(PQRS,\angle P + \angle R = {180^{\rm{o}}}\) and \(\angle S + \angle Q = {180^{\rm{o}}}.\)
Therefore, draw a circle passing through the points \(P,\, Q\) and \(R\) and observe that it also passes through the point \(S.\)
Hence, we conclude that quadrilateral \(PQRS\) is a cyclic quadrilateral.

Theorem 3: If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

To prove: \(\angle CBE = \angle ADC\)
Proof: Let the side \(AB\) of the cyclic quadrilateral \(ABCD\) is extended to \(E.\)
Here \(\angle ABC\) and \(\angle CBE\) are linear pair, their sum is \({180^{\rm{o}}}\) and the angles \(\angle ABC\) and \(\angle ADC\) are the opposite angles of a cyclic quadrilateral, and their sum is also \({180^{\rm{o}}}.\)
From this, \(\angle ABC + \angle CBE = \angle ABC + \angle ADC\) and finally we get \(\angle CBE = \angle ADC.\)
Similarly, we can prove it for all angles.

Solved Examples – Properties of Circle

Q.1. In the given figure, a circle with centre \(D\) touches the sides of \(\angle ACB\) at \(A\) and \(B.\) If \(\angle ACB = {52^{\rm{o}}},\) find the measure of \(\angle ADB.\)
Ans: We know that the sum of all angles of a quadrilateral is \({360^{\rm{o}}}.\)
\(\angle ACB\angle CAD + \angle CBD + \angle ADB = {360^{\rm{o}}}\)
\( \Rightarrow {52^{\rm{o}}} + {90^{\rm{o}}} + {90^{\rm{o}}} + \angle ADB = {360^{\rm{o}}}\) (By using the tangent theorem)
\( \Rightarrow \angle ADB + {232^{\rm{o}}} = {360^{\rm{o}}}\)
\(\therefore \angle ADB = {360^{\rm{o}}} – {232^{\rm{o}}} = {128^{\rm{o}}}\)

Q.2. The radius of the circle is \(14\,{\rm{cm,}}\) and the perpendicular distance from the chord to the centre is \(8\,{\rm{cm}}{\rm{.}}\) Identify the length of a chord?
Ans: Given radius, \(r = 14\,{\rm{cm}}\)
The perpendicular distance \(h = 8\,{\rm{cm}}\)
Now, the formula for the length of the chord\( = 2\sqrt {\left( {{r^2} – {h^2}} \right)} \)
So, the length of the chord \( = 2\sqrt {\left( {{{14}^2} – {8^2}} \right)} \)
\( = 2\sqrt {\left( {196 – 64} \right)} \)
\( = 2\sqrt {\left( {132} \right)} \)
\( = 2 \times 11.5\)
\( = 23\)
Hence, the length of a chord is \(23\,{\rm{cm}}\) approximately.

Q.3. \(ABCD\) is a cyclic quadrilateral in which \(AC\) and \(BD\) are its diagonals. If \(\angle DBC = {55^{\rm{o}}}\) and \(\angle BAC = {45^{\rm{o}}},\) find \(\angle BCD = {45^{\rm{o}}},\)
Ans: \(\angle CAD = \angle DBC = {55^{\rm{o}}}\) (Angles in the same segment are equal)
Therefore, \(\angle DAB = \angle CAD + \angle BAC = {55^{\rm{o}}} + {45^{\rm{o}}} = {100^{\rm{o}}}\)
But \(\angle DAB + \angle BCD = {180^{\rm{o}}}\) (Opposite angles of a cyclic quadrilateral)
So, \(\angle BCD = {180^{\rm{o}}} – {100^{\rm{o}}} = {80^{\rm{o}}}\)

Q.4. In the figure given below, find the value of \({x^{\rm{o}}}\) and \({y^{\rm{o}}}.\)
Ans: By the exterior angle property of a cyclic quadrilateral, if one side of a cyclic quadrilateral is produced, then the exterior angle equals the interior opposite angle.
So, we get, \({y^{\rm{o}}} = {100^{\rm{o}}}\)
and \({x^{\rm{o}}} + {30^{\rm{o}}} = {60^{\rm{o}}}\)
\( \Rightarrow {x^{\rm{o}}} = {60^{\rm{o}}} – {30^{\rm{o}}} = {30^{\rm{o}}}\)
So, \({x^{\rm{o}}} = {30^{\rm{o}}}\) and \({y^{\rm{o}}} = {100^{\rm{o}}}\)

Q.5. Two circles intersect at two points \(A\) and \(B.\,AD\) and \(AC\) are diameters to the two circles. Prove that \(B\) lies on the line segment \(DC.\)
Ans: Join \(AB.\)
\(\angle ABD = {90^{\rm{o}}}\) (Angle in a semicircle)
\(\angle ABC = {90^{\rm{o}}}\) (Angle in a semicircle)
So, \(\angle ABD + \angle ABC = {90^{\rm{o}}} + {90^{\rm{o}}} = {180^{\rm{o}}}\)
Therefore, \(DBC\) is a line. That is \(B\) lies on the line segment \(DC.\)

In this article, we have discussed the important properties of a chord of a circle, properties of a circle related to angles, properties of a circle related to tangent and properties of a circle related to cyclic quadrilaterals. The students can also check types of circles here.

With the help of proofs of theorems and solved examples related to the properties of a circle, one can easily understand the concept.

FAQs on Properties of Circle

Q.1. Can a cyclic quadrilateral be in a semicircle?
Ans: Yes, if we take the diameter as the one side of the quadrilateral and other sides lie on the circumference, then a cyclic quadrilateral can be in a semicircle.

Q.2. Is a square a cyclic quadrilateral?
Ans: Yes, the sum of the opposite angles of a square is supplementary. Hence, it is a cyclic quadrilateral.

Q.3. Does radius bisect a tangent?
Ans: No, it does not bisect the tangent, but it is perpendicular to the tangent through the point of contact.

Q.4. What are the chord properties of a circle?
Ans: The properties of a circle related to a chord are:
(i) A perpendicular from the centre of the circle to the chord bisects the chord.
(ii) The line drawn through the centre of the circle to bisect the chord is perpendicular to the chord.

Q.5. What is the measure of an angle on a diameter?
Ans: The measure of the angle on diameter is \({180^{\rm{o}}}.\)

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