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)?

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: 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 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: 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). In triangles OAM and OBM,

OA = O B                         (Why?)
AM = BM                        (Why?)   
OM = OM                     (Common)

Therefore, ΔOAM ΔOBM                    (How?)
This gives OMA = OMB = 90            (Why?)

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 (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 (ii)]. What will happen when you take three points A, B and C? Can you draw a circle passing through three collinear points?

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

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

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, 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, 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.