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# 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:

**Fig. 10.2**

*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.

**Fig. 10.3**

**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*.

**Fig. 10.4**

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.

**Fig. 10.5**

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*.

## 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.

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.

**Fig. 10.12**

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.

**Fig. 10.14**

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:

## 10.4 PERPENDICULARS FROM THE CENTRE TO A 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:

## 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).

**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:

Given an arc of a circle, complete the circle.

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).

**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 PL_{1}, PL_{2}, PM, PL_{3}, PL_{4}, 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.

**Fig. 10.21**

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.

**Fig. 10.22**

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:

**Fig. 10.23**

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

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.

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, Δ OLE Δ OME (Why ?)

This gives OL = OM (CPCT)

So, AB = CD (Why ?)

**Fig. 10.24**

## 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). This shows that equal chords make congruent arcs and conversely congruent arcs make equal chords of a circle. You can state it as follows:

**Fig. 10.26**

* If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.*

Also the angle subtended by an arc at the centre is defined to be angle subtended by the corresponding chord at the centre in the sense that the minor arc subtends the angle and the major arc subtends the reflex angle. Therefore, in Fig 10.27, the angle subtended by the minor arc PQ at O is ∠POQ and the angle subtended by the major arc PQ at O is reflex angle POQ.

In view of the property above and Theorem 10.1, the following result is true:

* Congruent arcs *(*or equal arcs*) *of a circle subtend equal angles at the centre.*

Therefore, the angle subtended by a chord of a circle at its centre is equal to the angle subtended by the corresponding (minor) arc at the centre. The following theorem gives the relationship between the angles subtended by an arc at the centre and at a point on the circle.

**Fig. 10.27**

**Remark : **Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP.

In Theorem 10.8, A can be any point on the remaining part of the circle. So if you take any other point C on the remaining part of the circle (see Fig. 10.29), you have

∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ

Therefore, ∠ PCQ = ∠ PAQ.

This proves the following:

**Fig. 10.29**

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 = ½ ∠POQ = ½ x 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:

## 10.8 CYCLIC QUADRILATERALS

A quadrilateral ABCD is called *cyclic *if all the four vertices of it lie on a circle (see Fig 10.31). You will find a peculiar property in such quadrilaterals. Draw several cyclic quadrilaterals of different sides and name each of these as ABCD. (This can be done by drawing several circles of different radii and taking four points on each of them.) Measure the opposite angles and write your observations in the following table.

**Fig. 10.31**

S.No. of Quadrilateral | ∠A | ∠ B | ∠ C | ∠ D | ∠ A +∠ C | ∠B +∠ D |
---|---|---|---|---|---|---|

1. | ||||||

2. | ||||||

3. | ||||||

4. | ||||||

5. | ||||||

6. |

What do you infer from the table?

You find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, neglecting the error in measurements. This verifies the following:

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

You can see the truth of this theorem by following a method similar to the method adopted for Theorem 10.10.

In Fig. 10.32, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that ∠AEB = 60°.

Join OC, OD and BC.

Triangle ODC is equilateral. (Why?)

Therefore, ∠ COD = 60°

Now, CBD =½ ∠ COD (Theorem 10.8)

This gives ∠ CBD = 30°

Again, ∠ ACB = 90° (Why ?)

So, ∠ BCE = 180° –∠ ACB = 90°

which gives ∠CEB = 90° – 30° = 60°, i.e. ∠AEB = 60°

**Fig. 10.32**

In Fig 10.33, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠ DBC = 55° and ∠BAC = 45°, find ∠BCD.

∠CAD = ∠ DBC = 55° (Angles in the same segment)

Therefore, ∠ DAB = ∠ CAD + ∠BAC

= 55° + 45° = 100°

But ∠ DAB + ∠ BCD = 180° (Opposite angles of a cyclic quadrilateral)

So, ∠ BCD = 180° – 100° = 80°

**Fig. 10.33**

Two circles intersect at two points A and B. AD and AC are diameters to the two circles (see Fig.10.34). Prove that B lies on the line segment DC.

Join AB.

∠ ABD = 90° (Angle in a semicircle)

∠ ABC = 90° (Angle in a semicircle)

So, ∠ ABD + ∠ ABC = 90° + 90° = 180°

Therefore, DBC is a line. That is B lies on the line segment DC.

**Fig. 10.34**

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

In Fig. 10.35, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH. Now, ∠ FEH = ∠AEB = 180° –∠ EAB –∠ EBA (Why ?)

= 180° –½(∠A + ∠B)

and ∠FGH = ∠CGD = 180° –∠GCD –∠GDC (Why ?)

= 180° –½ (∠ C + ∠D)

Therefore, ∠FEH + ∠FGH = 180° –½ (∠A + ∠ B) + 180° – ½ (∠ C + ∠ D)

= 360° –½ (∠ A+ ∠ B + ∠C +∠ D) = 360° – ½ x 360°

= 360° – 180° = 180°

Therefore, by Theorem 10.12, the quadrilateral EFGH is cyclic.

**Fig. 10.35**

In this chapter, you have studied the following points:

- A circle is the collection of all points in a plane, which are equidistant from a fixed point in the plane.
- Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
- If the angles subtended by two chords of a circle (or of congruent circles) at the centre (corresponding centres) are equal, the chords are equal.
- The perpendicular from the centre of a circle to a chord bisects the chord.
- The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
- There is one and only one circle passing through three non-collinear points.
- Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres).
- Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are equal.
- If two arcs of a circle are congruent, then their corresponding chords are equal and conversely if two chords of a circle are equal, then their corresponding arcs (minor, major) are congruent.
- Congruent arcs of a circle subtend equal angles at the centre.
- 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.
- Angles in the same segment of a circle are equal.
- Angle in a semicircle is a right angle.
- 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.
- The sum of either pair of opposite angles of a cyclic quadrilateral is 180
^{o}. - If sum of a pair of opposite angles of a quadrilateral is 180
^{o}, the quadrilateral is cyclic.