Chapter 17 - Circles

In the previous section we completed the discussion on irrational numbers. In this chapter, we will discuss about Circles.

We usually draw circles using the drawing tool - compass. Some times we will come across circles which have been drawn by bangles , large key rings, circular lids etc.,. In such cases, the centre of the circle will not be given. We will have to use mathematical methods to find it. Let us try:

Consider the green circle shown in fig.17.1(a) below. We need to find it's centre. For that, we will use one important property of the 'centre of any circle': The centre of a circle is at the same distance from any point on the circle.

Fig.17.1

Let us try to use this property. In the circle in fig.17.1(a), any two convenient points A and B are marked. Using the compass, draw an arc with A as centre. Using the same radius, draw another arc with B as centre, and cutting the first arc at C. Now, C is at the same distance from both A and B. But we can see that C is not the centre. It is placed towards the top side of the circle.

Also we can say this:

C is equidistant from Aand B. That is., AC = AB.

But what about any third point other than A and B on the circle?

Obviously, the distance from that third point to C, will not be equal to AC. So, C is not the centre. The distance AC that we took for the first arc, is not correct. Let us try another distance. See fig(b).

This time we take a smaller distance. The point of intersection of the two arcs is at C'. We can see that C' is also towards the top side. It is not the centre.

Take an even smaller distance. see fig(c). This time, the point of intersection of the two arcs is at C''. We can see that C'' is towards the bottom side. It is also not the centre.

We cannot continue like this using different trials. Even if we do find the centre after several trials, it may not be accurate. We need a different method. 

However, the trials that we did, do not go as waste. We find an important property from those trials:
• We find that, the trial points C, C' and C'', all lie on a straight line (drawn in cyan colour) as in fig(d).
• No matter how many trials that we do, all the points of intersection will lie on that line. 
■ With this information, we will think about a solution from a different point of view.

In the fig(d), we see pairs of equal sides: [AC and BC], [AC' and BC'] and [AC'' and BC''].

Do they resemble any thing that is already known to us? Does any thing come to mind?

Yes. You are correct. Isosceles triangles. Those triangles have two equal sides.

• What we have inside the circle, is a 'series of isosceles triangles'. All the triangles in the series have the same base: AB

• Fig.17.2(a) below shows such a series. PQ is the base for all the triangles in the series. R, R', and R'' are the apex points.

Fig.17.2

• All those apex points lie on the straight line RS

• Now, what is the peculiarity of this line RS?

• RS will be perpendicular to the base AB

• Also, RS will bisect the base AB

• In short, RS is the perpendicular bisector of base AB.

• This gives us a new information: The centre of the circle in fig.17.1(d) lies on the perpendicular bisector of AB. This is shown in the fig.17.2(b). The perpendicular bisector is named as XY. 
• But this information is not enough. The centre could be any where on XY. We need the precise position. For that, we think in this way:

• Initially, we took any two convenient points A and B. We arrived at the result that, the centre lies some where on the perpendicular bisector of AB. The same result should work for any other two points.

• So we take another two convenient points A' and B'. Then we draw the perpendicular bisector X'Y' of A'B'. This is shown in fig.17.2(c). Now we have:

    ♦ The centre should lie on XY

    ♦ The centre should lie on X'Y'

• There is only one point that will satisfy both the conditions. That point is the intersection of XY and X'Y'. So we mark the point of intersection as the centre. We successfully determined the centre point.

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We saw that the perpendicular bisector of the line joining any two points on the circle, passes through the centre of the circle. The ‘line joining any two points on the circle’ is called the chord. So we can write:

■ The perpendicular bisector of any chord of a circle, passes through the centre of the circle.

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So we determined the centre of the circle using the above property. Now we will see another application of the property:

Suppose we have only a part of the circle as shown in fig.17.3(a). We want to find the centre and draw the full circle.

Fig.17.3

• We can draw any two convenient chords inside the given portion, and find the centre [fig(b)].

• Once the centre is found, we can complete the full circle [fig(c)]. 

In many engineering drawings, we will see circles or 'portions of circles'. The centre will not be given. But the centre will be required to make modifications to the drawing. In such situations, we can use this method to find the centre.

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So the isosceles triangle helped us to derive the easy method to find the centre of any circle. Let us write the relationship between: • the base of an isosceles triangle, and • it’s apex point:

1. The perpendicular from the apex point bisects the base. 

The above statement can be written in two more ways:

2. The line joining the midpoint of the base and the apex is perpendicular to the base

3. The apex is on the perpendicular bisector of the base

The above three statements can be modified to suit any circle. We will write them in the form of a theorem.
Theorem 17.1:

1. The perpendicular from the centre of a circle to any of it’s chord, bisects the chord

2. The line joining the midpoint of the chord and the centre of the circle is perpendicular to the chord
3. The centre of the circle is on the perpendicular bisector of the chord

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In the next section, we will see more details about circles.

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