10. Circles

Class 9 Mathematics Chapter 10. Circles

Chapter 10. Circles

Important Note :

1. 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.

2. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.

3. A diameter is the longest chord and all diameters have the same length, which is equal to two times the radius.

4. The length of the complete circle is called its circumference.

5. A piece of a circle between two points is called an arc.

6. Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
7. 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.
8. The perpendicular from the centre of a circle to a chord bisects the chord.
9. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
10. There is one and only one circle passing through three non-collinear points.
11. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres).
12. Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are equal.
13. 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.
14. Congruent arcs of a circle subtend equal angles at the centre.
15. 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.

i.e.,
16. Angles in the same segment of a circle are equal.

i.e. ,
17. Angle in a semicircle is a right angle.
18. 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.
19. The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

i.e., and
20. If sum of a pair of opposite angles of a quadrilateral is 180° , the quadrilateral is cyclic.

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EXERCISE 10.1

1. Fill in the blanks :

(i) The centre of a circle lies in of the circle . (Exterior/interior)

Answer : interior .

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in of the circle . (exterior/interior)

Answer : Exterior .

(iii) The longest chord of a circle is a of the circle .

Answer : Diameter .

(iv) An arc is a when its ends are the ends of a diameter .

Answer : Semicircle .

(v) Segment of a circle is the region between an arc and of the circle .

Answer : The chord .

(vi) A circle divides the plane, on which it lies,in parts .

Answer : Three .

2. Write True or False : Given reasons for your answers .

(i) Line segment joining the centre to any point on the circle is a radius of the circle .

Answer : True .

(ii) A circle has only finite number of equal chords .

Answer : False .

(iii) If a circle is divided into three equal arcs, each is a major arc .

Answer : False .

(iv) A chord of a circle ,which is twice as long as its radius, is a diameter of the circles .

Answer : True .

(v) Sector is the region between the chord and its corresponding arc .

Answer : False .

(vi) A circle is a plane figure .

Answer : True .

EXERCISE 10.2

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Solution: Given, O and O’ be the centres of the two circles and AB = CD .

To prove : .

Proof: In figure :

In and , we have

[Equal chord]

[Radius]

[SSS]

[CPCT] Proved.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Solution: Given, O and O’ be the centres of the two circles and .

To prove : .

Proof: In figure :

In and , we have

[given]

[Radius]

[SAS]

[CPCT] Proved.

EXERCISE 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution: (i) There is no common point .

(ii) There is one common point .

(iii) There are two common points .

Therefore, the maximum number of common points is two .
2. Suppose you are given a circle. Give a construction to find its centre.

Solution: We construct a circle with centre O .

3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solutino: let O and O’ be the centres of the two circle intersecting at the points P and Q .

To prove : OO’ is the bisector of the chord PQ .

Construction: We join OP , OQ , O’P and O’Q .

Proof: In figure :

In and ,we have

[Radius]

[Common sides]

[SSS]

[CPCT]

In and ,we have

[Radius]

[Given]

[Common sides]

[SSS]

[CPCT]

Again,

and AP = AQ

Therefore, OO’ is the bisector of the chord PQ .

EXERCISE 10.4

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution: In figure :

Here, OM = 5 cm , OO’ = 4 cm and O’M = 3 cm

In , we have

So, OO’M is a right angled triangle at O’ .

i.e.,

Therefore, the length of the common chord is 8 cm .

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution: Given, O be a centre of a circle . PQ and RS are two equal chord intersecting at C .

To Prove : PC = CR and CS = CQ .

Construction: Join OC . We draw and

Proof: In figure :

We know that , the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord .

So, ,, and

Again,

In and ,we have

[equidistant from centre]

[Common sides]

[RHS]

[CPCT]

Subtracting (ii) from (i) , we get

Adding (i) and (ii) , we get

Proved.

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution: Given, O be a centre of a circle . PQ and RS are two equal chord intersecting at C .

To prove :

Construction: Join OC . We draw and

Proof: In figure :

We know that , the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord .

So, and

In and ,we have

[equidistant from centre]

[Common sides]

[RHS]

[CPCT]

Proved.

4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Solution: Given, a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.

Construction: Join OP .

To prove : AB = CD .

Proof: In given figure :

We know that, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord .

i.e.,

So, and

Again,

So, and

Subtracting (i) from (ii) , we get

Proved.
5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip ?

Solution: In figure :

Here, Reshma , Salma and Mandip represented by R, S and M respectively .

Let, and we join OL .

Given, and

In ∆ORS , we have

and

From (i) and (ii), we get

Again,

So,

Therefore, the distance between Reshma and Mandip is 12 m .

6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution: In figure :

Here, Ankur , Syed and David represented by A, S and D respectively .

Given, and

Let

and

In , we have

From (i) and (ii) , we get

Therefore, the length of the string of each phone is .

EXERCISE 10.5

1. In Fig. 10.36, A,B and C are three points on a circle with centre O such that and . If D is a point on the circle other than the arc ABC, find .

Solution: Given, and

We have,

Again,

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution: let O be a centre of a circle and BC is a chord .

Given,

So, BOC is an equilateral triangle .

In ,we have

So,

Again ,

3. In Fig. 10.37, , where P, Q and R are points on a circle with centre O. Find .

Solution: Given,

Since, OP = OR

So, POQ is an isosceles triangle .

In , we have

4. In Fig. 10.38, , find .

Solution: Given, and

In , we have

We know that angles in the same segment of a circle are equal .

5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that and . Find .

Solution: Given, and

We have,

In , we have

We know that angles in the same segment of a circle are equal .

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If is 30°, find . Further, if AB = BC, find .

Solution: Given, and

We know that angles in the same segment of a circle are equal .

[On BC segment]

So,

Again, [On DC segment]

In , we have

Since , AB = BC

So, ABC is an isosceles triangle .

We have,

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution: Let PR and QS are the two diagonals of a cyclic quadrilateral are diameters of the circle through the vertices P , Q , R and S respectively .

To Prove : PQRS is a rectangle .

Proof: We know that ,the angle in a semicircle is a right angle .

So, and [PQ is a diameter]

Again, and[QS is a diameter]

In and, we have

[diameter]

[common sides]

[RHS]

[CPCT]

Similarly , we show that

Therefore, PQRS is a rectangle .

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution: let ABCD is a trapezium and AD = BC .

To prove : ABCD is a cyclic quadrilateral .

Construction: We draw and .

Proof: In figure :

In and, we have

[given]

[Equal distance from two parallel lines]

[RHS]

[CPCT]

Also,

[ and]

Again,

We know that the sum of a pair of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic .

Therefore, ABCD is a cyclic quadrilateral .

9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that .

Solution: Given, two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively .

To prove : .
Proof: We know that , the angles in the same segment of a circle are equal .

[same segment AP] …….. (i)

and [same segment DQ] ……….(ii)

Again, [Vertically opposite angle] ……….(iii)

From (i) , (ii) and (iii) , we get

Proved.

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution: Given, two circle intersecting at two points C and D such that AC and BC are diameters to the two circles and also the sides of the triangle ABC respectively .

To prove that D lies on the line segment DC .

Construction: We join CD .

Proof : In figure ,

We have , [Angle in a semicircle]

and [Angle in a semicircle ]

So,

Therefore, ADB is a line .

Thus , D lies on the line segment AB .

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that .

Solution: Given, ABC and ADC are two right triangles with common hypotenuse AC.

To prove : .

Proof: ?In figure :

We have,

Since, the sum of a pair of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic .

So, ABCD is a cyclic quadrilateral .

Again, the angles in the same segment of a circle are equal .

So, [On the segment CD ]

Proved.

12. Prove that a cyclic parallelogram is a rectangle.

Solution: let PQRS is a cyclic parallelogram .

To prove : PQRS is a rectangle .

Proof: In figure ,

Since, PQRS is a parallelogram .

So, and

Again, PQRS is a cyclic parallelogram ,

So,

Similarly ,

Thus, PQRS is a rectangle . Proved.

EXERCISE 10.6 (Optional)*

1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are , and .

9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
10. In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.