CIRCLES

Theorem 10.1 : Equal chords of a circle subtend equal angles at the centre.

Proof : You are given two equal chords AB and CD

of a circle with centre O . You want

to prove that ∠ AOB = ∠ COD.

In triangles AOB and COD,

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

AB = CD (Given)

Therefore, ∆ AOB ≅ ∆ COD (SSS rule)

This gives ∠ AOB = ∠ COD

               (Corresponding parts of congruent triangles) „

Theorem 10.2 : If the angles subtended by the chords of a circle at the centre

are equal, then the chords are equal.

The above theorem is the converse of the Theorem 10.1.

if you take ∠ AOB = ∠ COD, then

           ∆ AOB ≅ ∆ COD (Why?)

Can you now see that AB = CD?

Theorem 10.2 : If the angles subtended by the chords of a circle at the centre

are equal, then the chords are equal.

The above theorem is the converse of the Theorem 10.1. 

if you take ∠ AOB = ∠ COD, then

           ∆ AOB ≅ ∆ COD (Why?)

Can you now see that AB = CD?

Theorem 10.3 : The perpendicular from the centre of a circle to a chord bisects

the chord.

What is the converse of this theorem? To write this, first let us be clear what is

assumed in Theorem 10.3 and what is proved. Given that the perpendicular from the

centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the

converse, what the hypothesis is ‘if a line from the centre bisects a chord of a

circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the

converse is:

Theorem 10.4 : The line drawn through the centre of a circle to bisect a chord is

perpendicular to the chord.

Is this true? Try it for few cases and see. You will

see that it is true for these cases. See if it is true, in

general, by doing the following exercise. We will write

the stages and you give the reasons.

Let AB be a chord of a circle with centre O and

O is joined to the mid-point M of AB. You have to

p r o v e   t h a t   O M   ⊥ A B .   J o i n   O A  a n d   O B

(see Fig. 10.16). In triangles OAM and OBM,

OA = OB (Why ?)

AM = BM (Why ?)

OM = OM (Common)

Therefore, ∆OAM  ≅ ∆OBM (How ?)

This gives ∠OMA = ∠OMB = 90°    (Why ?)

Theorem 10.5 : There is one and only one circle passing through three given

non-collinear points

Theorem 10.6 : Equal chords of a circle (or of congruent circles) are equidistant

from the centre  (or centres).

Next, it will be seen whether the converse of this theorem is true or not. For this,

draw a circle with centre O. From the centre O, draw two line segments OL and OM

of equal length and lying inside the circle  Then draw chords PQ

and RS of the circle perpendicular to OL and OM respectively 

Measure the lengths of PQ and RS. Are these different? No, both are equal. Repeat

the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 10.6 which is stated as follows:

Theorem 10.7 : Chords equidistant from the centre of a circle are equal in

length.

We now take an example to illustrate the use of the above results:

Theorem 10.8 : 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 an arc PQ of a circle subtending angles POQ at the centre O and

PAQ at a point A on the remaining part of the circle. We need to prove that

∠ POQ = 2 ∠ PAQ

Consider the three different cases as  In (i), arc PQ is minor; in (ii),

arc PQ is a semicircle and in (iii), arc PQ is major.

Let us begin by joining AO and extending it to a point B.

In all the cases,

∠ BOQ = ∠ OAQ + ∠ AQO

because an exterior angle of a triangle is equal to the sum of the two interior opposite

angles.

Theorem 10.9 : Angles in the same segment of a circle are equal.

Again let us discuss the case (ii) of Theorem 10.8 separately. Here ∠PAQ is an angle

in the segment, which is a semicircle. Also, ∠ PAQ = 1/2∠ POQ = 1/2× 180° = 90°.

If you take any other point C on the semicircle, again you get that

∠ PCQ = 90°

Therefore, you find another property of the circle as:

Angle in a semicircle is a right angle.

The converse of Theorem 10.9 is also true. It can be stated as:

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 line segment,

the four points lie on a circle  (i.e. they are concyclic).You can see the truth of this result as follows:

In Fig. 10.30, AB is a line segment, which subtends equal angles at two points C and

D. That is∠ ACB = ∠ ADB

To show that the points A, B, C and D lie on a circle

let us draw a circle through the points A, C and B.

Suppose it does not pass through the point D. Then it

will intersect AD (or extended AD) at a point, say E

(or E′).

If points A, C, E and B lie on a circle,

∠ ACB = ∠ AEB (Why?)

But it is given that ∠ ACB = ∠ ADB.

Therefore, ∠ AEB = ∠ ADB.

This is not possible unless E coincides with D. (Why?)

Similarly, E′ should also coincide with D.

10.8 Cyclic Quadrilaterals

A quadrilateral ABCD is called cyclic if all the four vertices

of it lie on a circle .You will find a peculiar

property in such quadrilaterals. Draw several cyclic

quadrilaterals of different sides and name each of these

as ABCD. 

T h e o r e m   1 0 . 11   : T h e   s u m   o f   e i t h e r   p a i r   o f   o p p o s i t e   a n g l e s   o f   a   c y c l i c

quadrilateral is  180º.

In fact, the converse of this theorem, which is stated below is also true.

Theorem 10.12 : If the sum of a pair of opposite angles of a quadrilateral is

180º, the quadrilateral is cyclic.

You can see the truth of this theorem by following a method similar to the method

adopted for Theorem 10.10.

FOR REFERENCE OF DIAGRAMS PLEASE REFER THE NCERT BOOK OF CLASS 9TH.....WE HAVE STATED ABOUT THE FIGURE.