**NCERT Class 9 Mathematics Chapter 9 Circles** **Solutions,** **NCERT Solutions For Class 9 Maths, NCERT Class 9 Mathematics Chapter 9 Circles Notes** to each chapter is provided in the list so that you can easily browse throughout different chapter **NCERT Class 9 Mathematics Chapter 9 Circles****Notes** and select needs one.

**NCERT Class 9 Mathematics Chapter 9 Circles**

**NCERT Class 9 Mathematics Chapter 9 Circles**Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. **NCERT****Class 9 Mathematics Textual Question Answer. **These solutions are part of NCERT All Subject Solutions. Here we have given

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**Solutions****NCERT Class 9 Mathematics Chapter 9 Circles****Circles**

**Circles****Chapter – 9**

Exercise 9.1 |

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

Ans: Given: AB and CD are two equal chords of congruent circles with centres O and O’ respectively.

To Prove: ∠AOB= ∠CO’D

Proof: In ∆AOB and ∆ O’CD,

OA = O’C | Radii of congruent circles

OB = O’D | Radii of congruent circles

AB = CD | Given

∴ ∆ OAB ≅ O’CD | SSS Rule

∴ ∠AOB ≅ ∠ CO’D | c.p.c.t.

∴ ∠AOB= ∠ CO’D

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

Ans: ∠ AOB and ∠ CO’D are the two equal angles subtended by the chords AB and CD of two congruent circles with centres O and O’ respec- tively.

To Prove: AB = CD

Prove: In ΔΟΑΒ and ∆O’CD,

OA = O’C | Radii of congruent circles

OB = O’D | Radii of congruent circles

∠AOB = ∠CO’D | Given

∴ ΔΟΑΒ ≅ O’CD | SAS Rule

∴ AB = CD | c.p.c.t.

**Q.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.**

Ans: We know that if two circles in-tersect each other at two points, then line joining their centres is the particular bisector of their common chord.

∴ Length of the common chord

= PQ = 2O’P

= 2 x 3cm = 6 cm

**Q.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.**

Ans: Given: A circle with centre O.

Its two equal chords AB and CD inter- sect at E.

To Prove: AE = DE and CE = BE

Construction: Draw ON⊥CD. Join OE

Proof: In ∆OME and ∆ONE,

OM = ON

| ∵ Equal chords of a circle are equidistant from the centre

OE = OE | Common

∴ ∆ OME ≅ ∆ONE | R.H.S.

∴ ME = NE | c.p.c.t.

⇒ AM + ME = DN + NE

⇒ AB – AE = CD – DE

⇒ BE = CE | Given

**Q.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.**

Ans: Given: Two equal chords AB and CD of a circle with centre O inter- sect within the circle. Their point of intersection is E.

To Prove: ∠OEA = ∠ OED

Construction: Join OA and OD.

Proof: In ∆OEA and ∆OED,

OE = OE | Common

OA = OD | Radii of a circle

AE = DE | Proved in Example 2 above

∴ ∆ OEA cong ∆ OED | SSS Rule

∴ ∠ OEA =∠OED | c.p.c

**Q.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 figure]**

Ans: Given: A line intersects two concentric circles (circles with the same centre) with cen- tre O at A, B, C and D.

To Prove: AB = CD

Construction: Draw OM⊥BC

Proof: The perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ AM=DM …(1)

and BM = CM …(2)

Subtracting (2) from (1), we get

AM-BM-DM-CM

⇒ AB = CD

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

Ans: In ∆NOR and ∆NOM,

ON=ON | Common

∠NOR = ΔΝΟΜ

|∵ Equal chords of a circle subtend equal angles at the centre

OR=OM | Radii of a circle

∴ ∆NOR≅∆NOM | SAS Rule

∴ ∠ONR = ∠ONM | c.p.c.t.

and NR = NM | c.p.c.t.

But ∠ONR+∠ONM=180⁰ | Linear Pair Axiom

∴ ∠ONR = ∠ONM = 90⁰

∴ ON is the perpendicular bisector of RM

Draw bisector SN of ∠RSM to intersect the chord RM in N.

In ∆RSM and ∆MSN,

RS = MS (= 6 cm each)

SN=SN | Common

∠RSN = ∠MSN | By construction

∴ ∆RSN≅AMSN | SAS Rule

∴ ∠RNS=∠MNS | c.p.c.t.

and RN = MN | c.p.c.t.

But ∠RNS +∠MNS = 180⁰ | Linear Pair Axiom

∴ ∠RNS = ∠MNS = 90⁰

∴ SN is the perpendicular bisector of RM and therefore passes through O when produced.

Let ON = xm

Then SN = (5 – x)m

In right triangle ONR,

x²+ RN² = 5² …(1)

| By Pythagoras Theorem In right triangle SNR,

(5 – x)² + RN² = 6² …(2)

| By Pythagoras Theorem

From (1),

RN² = 5²- x²

From (2),

RN² = 6²- (5 – x)²

Equating the two values of RN² ,we get

5² – x² = 6² – (5 – x)²

⇒ 25-x² = 36-(25-10x+x)²⇒ 25 – x² = 36 – 25 + 10x – x²

⇒ 25 – x² = 11+10x-x²⇒ 25-11=10x

⇒14 = 10x ⇒10 x = 14

Put x = 1.4 in (1), we get

(1.4)² + RN² = 5²

⇒ RN² = 5² – (1.4)²

⇒ RN² = 25 – 1.96

⇒ RN² = 23.04

⇒ RN = √(23.04)

⇒ RN = 4.8

∴ RM = 2RN = 2×4.8m = 9.6m

Hence the distance between Reshma and Mandip is 9.6 m.

**Q.6. A circular part 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 to each other. Find the length of the string of each phone.**

Ans: Length of the string of each phone = Given equal distance.

Exercise 9.2 |

**Q.1. In figure, A, B and C are three points on a circle with centre O such that angle BOC = 30⁰ and ∠ AOB = 60⁰ If D is a point on the circle other than the arc ABC, find ∠ADC.**

Ans:

| The angle subtended by an the angle subtended by it at any point on the remaining part of the circle

**Q.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.**

Ans: ∵ OA = OB = AB | Given

∴ ∆AOB is equilateral

∴ ∠ AOB = 60⁰

| 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

Now, ∵ ADBC is a cyclic quadrilateral.

∴ ∠ADB + ∠ACB = 180⁰

| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰

⇒∠ADB + 30⁰= 180⁰

⇒ ∠ ADB = 180⁰- 30⁰

⇒ ∠ADB = 150⁰

**Q.3. In figure, ∠ PQR = 100⁰, where P, Q and R are points on a circle with centre O. Find ∠OPR.**

Ans: Take a point S in the major arc. Join PS and RS.

∵ PQRS is a cyclic quadrilateral.

∵ ∠ PQR+ ∠PSQ = 180⁰

| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰

⇒ 100⁰ + ∠PSR = 180⁰

⇒ ∠ PSR = 180⁰ – 100⁰

⇒∠PSR = 80⁰ …(1)

Now, ∠POR = 2∠PSR = 2 × 80⁰ = 160⁰ … (2) | Using (1)

| 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

In, DOPR

∵ OP=OR | Radii of a circle

∴ ∠OPR = ∠ORP …(3)

| Angles opposite to equal sides of a triangle are equal In ∆OPR,

∠OPR + ∠ORP + ∠POR = 180⁰

Sum of all the angles of a triangle is 180⁰

⇒ ∠OPR + ∠OPR + 160 = 180⁰ | Using (2) and (1)

⇒ 2 ∠OPR + 160⁰ = 180⁰

⇒ 2 ∠OPR = 180⁰ – 160⁰ = 20⁰

⇒ ∠ OPR = 10⁰

**Q.4. In figure,**

∠ ABC = 69⁰,∠ACB = 31⁰ find ∠BDC

Ans: In ∆ ABC,

| Sum of all the angles of a triangle is 180⁰

⇒ ∠ BAC +∠ABC + ∠ ACB = 180⁰

⇒ ∠ BAC + 100⁰ = 180⁰

⇒ ∠BAC = 180⁰- 100⁰= 80⁰

Now, ∠ BDC = ∠ BAC = 80⁰

| Angles in the same segment of a circle are equal

**Q.5. In figure, A, B, C and D are four points on a circle. AC and BD in- tersect at a point E such that ∠ BEC = 130⁰ and ∠ BCD = 20⁰ Find ∠ BAC.**

Ans: ∠CED+ ∠ BEC = 180⁰ | Linear Pair Axiom

⇒∠CED + 130⁰ = 180⁰

⇒∠CED = 180⁰- 130⁰ = 50⁰ …(1)

∠ ECD = 20⁰ …(2)

In ∆CED,

∠CED + ∠ECD + ∠CDE = 180⁰

| Sum of all the angles of a triangle is 180⁰

⇒ 50⁰+20⁰ +∠CDE = 180⁰ | Using (1) and (2)

⇒ 70⁰+∠CDE = 180⁰

⇒ ∠CDE = 180⁰ – 130⁰ = 50⁰ …(1)

∠ ECD = 20⁰ …(2)

In ∆CED,

∠CED + ∠ECD + ∠CDE = 180⁰

| Sum of all the angles of a triangle is 180⁰

⇒ 50⁰ + 20⁰ +∠CDE = 180⁰ | Using (1) and (2)

⇒ 70⁰ +∠CDE = 180⁰

⇒ ∠CDE = 180⁰ – 70⁰

⇒∠CDE = 110⁰

Now, ∠BAC= ∠CDE | Angles in the same segment of a circle are equal

= 110⁰ | Using (3)

**Q.6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If angle DBC = 70⁰, ∠BAC is 30⁰, find ∠BCD. Further, if AB = BC, find ∠ECD.**

Ans: ∠CDB=∠BAC | Angles in the same segment of a circle are equal

= 30° …(1)

∠ DBC = 70⁰ …(2)

In ∆BCD,

∠BCD + ∠BDC + ∠CDB = 180⁰

| Sum of the angles of triangle is 180⁰

⇒ ∠BCD + 70⁰+ 30⁰ = 180⁰ | Using (1) and (2)

⇒ ∠BCD + 100⁰ = 180⁰

⇒ ∠BCD = 180⁰ – 100⁰

⇒ ∠BCD = 80⁰ …(3)

In ∆ABC, AB = BC

∴ ∠BCA = ∠BAC | Angles opposite to equal sides of A a triangle are equal

= 30⁰ …(4) | ∵ ∠ BAC=30⁰ (given)

Now, ∠ BCD = 80⁰ | From (3)

⇒ ∠BCA+ ∠ECD = 80⁰

⇒ 30⁰ + ∠ECD = 80⁰

⇒ ∠ECD = 80⁰ – 30⁰

⇒ ∠ECD = 50⁰

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

Ans: In and ∆OCD,

OA = OC | Radii of a circle

OB = OD | Radii of a circle

∠AOB= ∠COD | Vertically Opposite Angles

∴ ΔΑΟΒ ≅ ∆OCD | SAS Rule

∴ AB = CD | c.p.c.t.

⇒ Arc AB = Arc CD …(1)

Similarly, we can show that

Arc AD=Arc CB …(2)

Adding (1) and (2), we get

Arc AB+ Arc AD + Arc CB

⇒ Arc BAD Arc BCD

⇒ BD divides the circle into two equal parts (each semicircle)

∴ ∠ A = 90⁰ ∠C = 90⁰ | Angle of a semicircle is 90⁰

Similarly, we can show that

∠B = 90⁰, ∠D=90⁰

∴ ∠A= ∠B= ∠C= ∠D = 90⁰

∴ ABCD is a rectangle.

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

Ans: Given: ABCD is cyclic.

Construction: Draw BE || AD

Proof: ∵ AB || DE | Given

AB || BE | By construction

∴ Quadrilateral ABCD is a parallelogram.

∴ ∠BAD = ∠BED …(1) | Opp. ∠s of all || gm

and AD = BE …(2) | Opp. sides of a || gm

But AD = BC …(3) | Given

From (2) and (3), BE = BC

∴ ∠BEC = ∠BCE …(4) | Angles opposite to equal sides

∠BEC+ ∠BED = 180⁰ | Linear Pair Axiom

⇒ ∠BCE+∠BAD = 180⁰ | From (4) and (1)

⇒ Trapezium ABCD is cyclic.

∵ If a pair of opposite angles of a quadrilateral is 180⁰, then the quadrilateral is cyclic.

**Q.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 figure). Prove that ∠ACP = ∠QCD.**

Ans: Given: Two circles in- tersect at two points B and C.

Through B. two line segments ABD and PBQ are drawn to in- tersect the circles at A, D and P,Q respectively.

To Prove: ∠ACP = ∠QCD

Proof: ∠ACP = ∠ABP…(1) | Angles in the same segment of a circle are equal

∠QCD = ∠QBD…(2) | Angles in the same seg- ment of a circle are equal

∠ABD = ∠QBD…(3) | Vertically Opposite Angles

From (1), (2) and (3), ∠ACP = ∠QCD

**Q.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.**

Ans: Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect in a point D.

To Prove: D lies on the third side BC of ∆АВС.

Construction: Join AD.

Proof: ∵ Circle described on AB as diameter intersects BC in D.

∴ ∠ADB=90⁰ | Angle in a semi-circle

But ∠ADB + ∠ADC = 180⁰ | Linear Pair Axiom

∴ ∠ADC=90⁰

Hence, the circle described on AC as diameter must pass through D.

Thus, the two circles intersect in D.

Now, ∠ADC=180⁰.

∴ Points B, D, C are collinear.

∴ D lies on BC.

**Q.11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD**

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

To Prove: ∠CAD = ∠CBD

Proof: ∵ AC is the common hypotenuse and ABC and ADC are two right triangles.

∴ ∠ABC = 90⁰ = ∠ADC

⇒ Both the triangles are in the same semi-circle.

∴ Points A, B, D and C are concyclic.

DC is a chord

∴ ∠CAD = ∠CBD | Angles in the same segment are equal

**Q.12. Prove that a cyclic parallelogram is a rectangle.**

Ans: Given: ABCD is a cyclic parallelogram.

To Prove: ABCD is a rectangle.

Proof: ∵ ABCD is a cyclic quadrilateral

∴ ∠1+ ∠2 = 180⁰ …(1)

| ∵ Opposite angles of a cyclic quadrilateral are supplementary

∴ ABCD is a parallelogram

∴ ∠1+ ∠2 = …(2) | Opp. angles of a || gm

From (1) and (2), ∠1 = ∠2 = 90⁰

∴ || gm ABCD is a rectangle.