• Perpendicular from the Centre to a Chord

• Circle through Three Points

• Circle through Three Points

`color{blue}("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:

`color{red}("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:

`color{navy}("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,

`color{navy}(OA = OB \ \ \ \ \) `

`color{red}(AM = BM \ \ \ \ \ \ \)`

`color{green}(OM = OM\ \ \ ("Common"))`

`color{orange}("Therefore," DOAM ≅ DOBM\ \ )`

`color{orange}("This gives" ∠OMA = ∠OMB = 90° \ \ \ \ )`

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:

`color{red}("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:

`color{navy}("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,

`color{navy}(OA = OB \ \ \ \ \) `

`color{red}(AM = BM \ \ \ \ \ \ \)`

`color{green}(OM = OM\ \ \ ("Common"))`

`color{orange}("Therefore," DOAM ≅ DOBM\ \ )`

`color{orange}("This gives" ∠OMA = ∠OMB = 90° \ \ \ \ )`

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?

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)].

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:

`color{orange}("Theorem 10.5 :")"There is one and only one circle passing through three given non-collinear points."`

`color{orange}("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 `Delta` ABC.

Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

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?

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)].

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:

`color{orange}("Theorem 10.5 :")"There is one and only one circle passing through three given non-collinear points."`

`color{orange}("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 `Delta` ABC.

Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Q 3210378219

Given an arc of a circle, complete the circle.

Class 9 Chapter 10 Example 1

Class 9 Chapter 10 Example 1

Let are 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).