Chapter 10

Circles

10.1 Introduction

You may have come across many objects in daily life, which are round in shape, such as wheels of a vehicle, bangles, dials of many clocks, coins of denominations 50 p, Re 1 and Rs 5, key rings, buttons of shirts, etc. (see Fig.10.1). In a clock, you might have observed that the second’s hand goes round the dial of the clock rapidly and its tip moves in a round path. This path traced by the tip of the second’s hand is called a circle. In this chapter, you will study about circles, other related terms and some properties of a circle.

Fig. 10.1

10.2 Circles and Its Related Terms: A Review

Take a compass and fix a pencil in it. Put its pointed leg on a point on a sheet of a paper. Open the other leg to some distance. Keeping the pointed leg on the same point, rotate the other leg through one revolution. What is the closed figure traced by the pencil on paper? 

As you know, it is a circle (see Fig.10.2).How did you get a circle? You kept one point fixed (A in Fig.10.2) and drew all the points that were at a fixed distance from A. This gives us the following definition:

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle. In Fig.10.3, O is the centre and the length OP is the radius of the circle.

Remark : Note that the line segment joining the centre and any point on the circle is also called a radius of the circle. That is, ‘radius’ is used in two senses-in the sense of a line segment and also in the sense of its length.

You are already familiar with some of the following concepts from Class VI. We are just recalling them.

A circle divides the plane on which it lies into three parts. They are: (i) inside the circle, which is also called the interior of the circle; (ii) the circle and (iii) outside the circle, which is also called the exterior of the circle (see Fig.10.4).The circle and its interior make up the circular region.

If you take two points P and Q on a circle, then the line segment PQ is called a chord of the circle (see Fig. 10.5). The chord, which passes through the centre of the circle, is called a diameter of the circle. As in the case of radius, the word ‘diameter’ is also used in two senses, that is, as a line segment and also as its length. Do you find any other chord of the circle longer than a diameter? No, you see that a diameter is the longest chord and all diameters have the same length, which is equal to two times the radius. In Fig.10.5, AOB is a diameter of the circle. How many diameters does a circle have? Draw a circle and see how many diameters you can find.

A piece of a circle between two points is called an arc. Look at the pieces of the circle between two points P and Q in Fig.10.6.

You find that there are two pieces, one longer and the other smaller
(see Fig.10.7).

The longer one is called the major arc PQ and the shorter one is called the minor arc PQ. The minor arc PQ is also denoted by and the major arc PQ by , where R is some point on the arc between P and Q. Unless otherwise stated, arc PQ or stands for minor arc PQ. When P and Q are ends of a diameter, then both arcs are equal and each is called a semicircle.

The length of the complete circle is called its circumference. The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. You will find that there are two types of segments also, which are the major segment and the minor segment (see Fig. 10.8). The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector. Like segments, you find that the minor arc corresponds to the minor sector and the major arc corresponds to the major sector. In Fig. 10.9, the region OPQ is the minor sector and remaining part of the circular region is the major sector. When two arcs are equal, that is, each is a semicircle, then both segments and both sectors become the same and each is known as a semicircular region.

Fig. 10.8

Fig. 10.9

10.3 Angle Subtended by a Chord at a Point

Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR (see Fig. 10.10). Then ∠ PRQ is called the angle subtended by the line segment PQ at the point R. What are angles POQ, PRQ and PSQ called in Fig. 10.11? ∠ POQ is the angle subtended by the chord PQ at the centre O, ∠ PRQ and ∠ PSQ are respectively the angles subtended by PQ at points R and S on the major and minor arcs PQ.

Fig. 10.10

Fig. 10.11

Let us examine the relationship between the size of the chord and the angle subtended by it at the centre. You may see by drawing different chords of a circle and

angles subtended by them at the centre that the longer is the chord, the bigger will be the angle subtended by it at the centre. What will happen if you take two equal chords of a circle? Will the angles subtended at the centre be the same or not?

Draw two or more equal chords of a circle and measure the angles subtended by them at the centre (see Fig.10.12).

You will find that the angles subtended by them at the centre are equal. Let us give a proof of this fact.

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 (see Fig.10.13).

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)

Remark : For convenience, the abbreviation CPCT will be used in place of ‘Corresponding parts of congruent triangles’, because we use this very frequently as you will see.

Now if two chords of a circle subtend equal angles at the centre, what can you say about the chords? Are they equal or not? Let us examine this by the following activity:

Take a tracing paper and trace a circle on it. Cut it along the circle to get a disc. At its centre O, draw an angle AOB where A, B are points on the circle. Make another angle POQ at the centre equal to ∠ AOB. Cut the disc along AB and PQ (see Fig. 10.14).

You will get two segments ACB and PRQ of the circle. If you put one on the other, what do you observe? They cover each other, i.e., they are congruent. So AB = PQ.

Though you have seen it for this particular case, try it out for other equal angles too. The chords will all turn out to be equal because of the following theorem:

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. Note that in Fig. 10.13, if you take ∠ AOB = ∠ COD, then

∆ AOB ≅ ∆COD (Why?)

Can you now see that AB = CD?

10.4 Perpendicular from the Centre to a Chord

Activity : Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. Does the point B coincide with A (see Fig.10.15)?

Yes it will. So MA = MB.

Give a proof yourself by joining OA and OB and proving the right triangles OMA and OMB to be congruent. This example is a particular instance of the following result:

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 prove that OM ⊥ AB. Join OA and OB
(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 ?)

10.5 Circle through Three Points

You have learnt in Chapter 6, that two points are sufficient to determine a line. That is, there is one and only one line passing through two points. A natural question arises. How many points are sufficient to determine a circle?

Take a point P. How many circles can be drawn through this point? You see that there may be as many circles as you like passing through this point [see Fig. 10.17(i)]. Now take two points P and Q. You again see that there may be an infinite number of circles passing through P and Q [see Fig.10.17(ii)]. What will happen when you take three points A, B and C? Can you draw a circle passing through three collinear points?

Fig. 10. 17

No. If the points lie on a line, then the third point will lie inside or outside the circle passing through two points (see Fig 10.18).

So, let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [see Fig. 10.19(i)]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicular bisectors intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) [see Fig. 10.19(ii)].

Fig. 10.19

Now O lies on the perpendicular bisector PQ of AB, you have OA = OB, as every point on the perpendicular bisector of a line segment is equidistant from its end points, proved in Chapter 7.

Similarly, as O lies on the perpendicular bisector RS of BC, you get

OB = OC

So OA = OB = OC, which means that the points A, B and C are at equal distances from the point O. So if you draw a circle with centre O and radius OA, it will also pass through B and C. This shows that there is a circle passing through the three points A, B and C. You know that two lines (perpendicular bisectors) can intersect at only one point, so you can draw only one circle with radius OA. In other words, there is a unique circle passing through A, B and C. You have now proved the following theorem:

Theorem 10.5 : There is one and only one circle passing through three given non-collinear points.

Remark : If ABC is a triangle, then by Theorem 10.5, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the ∆ ABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Example 1 : Given an arc of a circle, complete the circle.

Solution : Let arc PQ of a circle be given. We have to complete the circle, which means that we have to find its centre and radius. Take a point R on the arc. Join PR and RQ. Use the construction that has been used in proving Theorem 10.5, to find the centre and radius.

Taking the centre and the radius so obtained, we can complete the circle (see Fig. 10.20).

10.6 Equal Chords and Their Distances from the Centre

Let AB be a line and P be a point. Since there are infinite numbers of points on a line, if you join these points to P, you will get infinitely many line segments PL₁, PL₂, PM, PL₃, PL₄, etc. Which of these is the distance of AB from P? You may think a while and get the answer. Out of these line segments, the perpendicular from P to AB, namely PM in Fig. 10.21, will be the least. In Mathematics, we define this least length PM to be the distance of AB from P. So you may say that:

The length of the perpendicular from a point to a line is the distance of the line from the point.

Note that if the point lies on the line, the distance of the line from the point is zero.

A circle can have infinitely many chords. You may observe by drawing chords of a circle that longer chord is nearer to the centre than the smaller chord. You may observe it by drawing several chords of a circle of different lengths and measuring their distances from the centre. What is the distance of the diameter, which is the longest chord from the centre? Since the centre lies on it, the distance is zero. Do you think that there is some relationship between the length of chords and their distances from the centre? Let us see if this is so.

Activity : Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O. Fold the figure so that D falls on B and C falls on A [see Fig.10.22 (i)]. You may observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat the activity by drawing congruent circles with centres O and O′ and taking equal chords AB and CD one on each. Draw perpendiculars OM and O′N on them [see Fig. 10.22(ii)].Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that O coincides with O′ and M coincides with N. In this way you verified the following:

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 [see Fig. 10.23(i)]. Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig 10.23(ii)]. 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:

Example 2 : If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Solution : Given that AB and CD are two chords of a circle, with centre O intersecting at a point E. PQ is a diameter through E, such that ∠ AEQ = ∠ DEQ (see Fig.10.24). You have to prove that AB = CD. Draw perpendiculars OL and OM on chords AB and

CD, respectively. Now

∠ LOE = 180° – 90° – ∠ LEO = 90° – ∠ LEO

(Angle sum property of a triangle)

= 90° – ∠ AEQ = 90° – ∠ DEQ

= 90° – ∠ MEO = ∠ MOE

In triangles OLE and OME,

∠ LEO = ∠ MEO (Why ?)

∠ LOE = ∠ MOE (Proved above)

EO = EO (Common)

Therefore, ∆OLE ≅ ∆OME (Why ?)

This gives OL = OM (CPCT)

So, AB = CD (Why ?)

10.7 Angle Subtended by an Arc of a Circle

You have seen that the end points of a chord other than diameter of a circle cuts it into two arcs – one major and other minor. If you take two equal chords, what can you say about the size of arcs? Is one arc made by first chord equal to the corresponding arc made by another chord? In fact, they are more than just equal in length. They are congruent in the sense that if one arc is put on the other, without bending or twisting, one superimposes the other completely.

You can verify this fact by cutting the arc, corresponding to the chord CD from the circle along CD and put it on the corresponding arc made by equal chord AB. You will find that the arc CD superimpose the arc AB completely (see Fig. 10.26).