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Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 10 |

Chapter Name | Circles |

Category | NCERT Exemplar |

## NCERT Exemplar Class 9 Maths Solutions Circles

**Exercise 10.1: Multiple Choice Questions (MCQs)**

**Question 1:****AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is****(a)** 17 cm ** (b)** 15 cm ** (c)** 4 cm **(d)** 8 cm**Solution:****(d)** Given, AD = 34 cm and AB = 30 cm

In figure, draw OL ⊥ AB.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

**Question 2:****In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to****(a)** 2 cm ** (b)** 3 cm**(c)** 4 cm **(d)** 5 cm**Solution:****(a)** We know that, the perpendicular from the centre of a circle to a chord bisects the chord.

AC = CB = ½ AB = ½ x 8 = 4 cm

given OA = 5 cm

AO^{2} = AC^{2} + OC^{2 }

(5)^{2} = (4)^{2} + OC^{2}

25 = 16 + OC^{2}

OC^{2} = 25-16 = 9

OC = 3 cm

[taking positive square root, because length is always positive]

OA = OD [same radius of a circle]

OD = 5 cm

CD = OD – OC = 5 – 3 = 2 cm

**Question 3:****If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is****(a)** 6 cm ** (b)** 8 cm ** (c) 1**0 cm ** (d)** 12 cm**Solution:**

**Question 4:****In figure, if ∠ABC = 20°, then ∠AOC is equal to****(a)**20° ** (b)** 40° ** (c) **60° **(d)**10°**Thinking Process**

Use the theorem, that in a circale the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle and further simplify it.**Solution:****(b)** Given, ∠ABC = 20°

We know that, angle subtended at the centre by an arc is twice the angle subtended by it at the remaining part of circle.

∠AOC = 2∠ABC = 2 x 20° = 40°

**Question 5:****In figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to****(a)** 30° ** (b)** 60°**(c)** 90° ** (d)** 45°**Solution:**

**Question 6:****In figure, if ∠OAB = 40°, then ∠ACB is equal to From Eq. (i) ∠ACB = ∠ADB = 70°****(a)** 50° **(b)** 40° **(c)** 60° **(d)** 70°**Solution:****(a)**

In ΔQAB, OA = OB [both are the radius of a circle]

∠OAB = ∠OBA => ∠OBA = 40°

[angles opposite to equal sides are equal] Also, ∠AOB + ∠OBA + ∠BAO = 180°

[by angle sum property of a triangle]

∠AOB + 40° + 40° = 180°

=> ∠AOB = 180° – 80° = 100°

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

∠AOB = 2 ∠ACB => 100° =2 ∠ACB

∠ACB = 100°/2 = 50°

**Question 7:****In figure, if ∠DAB = 60° , ∠ABD = 50°, then ∠ACB is equal to****(a)** 60° ** (b)** 50° **(c)**70° **(d)** 80°**Thinking Process**

Use the theorem that angles in the same segment of a circle are equal and further simplify it.**Solution:****(c)** Given, ∠DAB = 60°, ∠ABD = 50°

Since, ∠ADB = ∠ACB …**(i)**

[**angles in same segment of a circle are equal**]

In ΔABD, ∠ABD + ∠ADB + ∠DAB = 180° [by angle sum property of a triangle]

50°+ ∠ADB + 60° = 180°

=> ∠ADB = 180° – 110° = 70°

**Question 8:****ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to****(a)** 80° ** (b)** 50° **(c)** 40° **(d)** 30°**Solution:****(b)** Given, ABCD is a cyclic quadrilateral and ∠ADC = 140°.

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

**Question 9:****In figure, BC is a diameter of the circle and ∠BAO = 60°. Then, ∠ADC is equal to OA = OB [both are the radius of circle]****(a)** 30° **(d)** 45° ** (d)** 60° **(d)** 120°**Solution:****(c)** In ΔAOB,

∠OBA = ∠BAO

[angles opposite to equal sides are equal]

∠OBA = 60° [∴ ∠BAO = 60°, given]

∠ABC=∠ADC

[angles in the same segment AC are equal]

∠ADC = 60°

**Question 10:****In figure, if ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to****(a)** 30° **(b)** 45° ** (c)** 90° ** (d)** 60°**Solution:****(d)**

In AOAB, ∠OAB + ∠ABO + ∠BOA = 180°

∠OAB + ∠OAB + 90° = 180° => 2∠OAB = 180°- 90°

[angles opposite to equal sides are equal] [angle sum property of a triangle] [**from Eq. (i)**]

=> ∠OAB = 90°/2 = 45° …**(i)**

In ΔACB, ∠ACB + ∠CBA + ∠CAB = 180°

∴ 45°+ 30°+ ∠CAB = 180°

=> ∠CAB = 180° – 75° = 105°

∠CAO+ ∠OAB = 105°

∠CAO + 45° = 105°

∠CAO = 105° – 45° = 60°

**Exercise 10.2: Very Short Answer Type Questions**

**Write whether True or False and justify your answer****Question 1:**

Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then, AB = CD.**Solution:****True**

Because, the chords equidistant from the centre of circle are equal in length.

**Question 2:**

Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then, ∠OAB = ∠OAC.**Solution:****False**

In figure, AB and AC are two chords of a circle. Join OB and OC.

In ΔOAB and ΔOAC,

**Question 3:**

The congruent circles with centres O and O’ intersect at two points A and B. Then, ∠AOB = ∠AO’B.**Solution:****True**

Join AB, OA and OB, O’A and BO’.

In ΔAOB and ΔAO’B,

OA = AO’ [both circles have same radius]

OB = BO’ [both circles have same radius]

and AB= AB [common chord]

ΔAOB = ΔAO’B [by SSC congruence rule]

=> ∠AOB = ∠AO’B [by CPCT]

**Question 4:**

Through three collinear points a circle can be drawn.**Solution:****False**

Because, circle can pass through only two collinear points but not through three collinear points.

**Question 5:**

A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.**Solution:****True**

Suppose, we consider diameter of a circle is AB = 66m.

AB

Then, radius of a circle = AB/2 = 6/2 = 3 cm, which is true.

**Question 6:**

If AOB is a diameter of a circle and C is a point on the circle, then AC^{2}+ BC^{2} = AB^{2}.**Solution:****True**

Since, any diameter of the circle subtends a right angle to any point on the circle.

If AOB is a diameter of a circle and C is a point on the circle, then ΔACB is right angled at C. In right angled ΔACB, [use Pythagoras theorem]

AC^{2} + BC^{2} = AB^{2}

**Question 7:**

ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.**Solution:****False**

In a cyclic quadrilateral, the sum of opposite angles is 180°.

Now, ∠A + ∠C = 90° + 95° = 185° ≠ 180°

and ∠B+∠D = 70° + 105° = 175° ≠ 180°

Here, we see that, the sum of opposite angles is not equal to 180°. So, it is not a cyclic quadrilateral.

**Question 8:**

If A, B, C and D are four points such that ∠BAC = 30° and ∠BDC = 60°, then D is the centre of the circle through A, B and C.**Solution:****False**

Because, there can be many points D, such that ∠BDC = 60° and each such point cannot be the centre of the circle through A, B and C.

**Question 9:**

If A, B, C and D are four points such that ∠BAC = 45° and ∠BDC = 45°, then A, B, C and D are concyclic.**Solution:****True**

Since, ∠BAC = 45° and ∠BDC = 45°

As we know, angles in the same segment of a circle are equal. Hence, A, B, C and D are concyclic.

**Question 10:****In figure, if AOB is a diameter and ∠ADC = 120°, then ∠CAB = 30°.****Solution:****True**

Join CA and CB.

Since, ADCB is a cyclic quadrilateral.

∠ADC + ∠CBA = 180°.

[sum of opposite angles of cyclic quadrilateral is 180°]

=>∠CBA = 180° -120° = 60° [**∴ ∠ADC = 120°**]

In ΔACB, ∠CAB + ∠CBA + ∠ACB = 180°

[by angle sum property of a triangle]

∠CAB + 60°+ 90°= 180°

[triangle formed from diameter to the circle is 90° i.e., ∠ACB = 90°)

=> ∠CAB = 180° – 150° = 30°.

**Exercise 10.3: Short Answer Type Questions**

**Question 1:**

If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.**Solution:**

Let AXB and CYD are arcs of circle whose centre and radius are O and r units, respectively.

Hence, the ratio of AB and CD is 1:1.

**Question 2:**

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB.

Thinking Process

Firstly, prove that ΔAPM is congruent to ΔBPM by SAS rule, then further prove the required result by CPCT rule.**Solution:**

Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O.

To prove arc PXA ≅ arc PYB

Construction Join AP and BP.

Proof In ΔAPM and ΔBPM,

AM = MB

∠PMA = ∠PMB

PM = PM

∴ ΔAPM s ΔBPM

∴PA = PB

=> arc PXA ≅ arc PYB

**Question 3:****A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.****Thinking Process**

- Firstly, inscribe a ΔABC in a circle, then draw the perpendicular bisecters of any two sides of a triangle.
- Secondly, prove that ΔOEA and ΔOEB are congruent by SAS rule and also ΔOMB and ΔOMC are congruent by RHS rule. Further, prove the required result.

**Solution:**

**Question 4:**

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.**Solution:**

Given AS and AC are two equal chords whose centre is O.

To prove Centre O lies on the bisector of ∠BAC.

Construction Join SC, draw bisector AD of ∠BAC.

Proof In ΔSAM and ΔCAM,

AS = AC [given]

∠BAM = ∠CAM [given]

**Question 5:**

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.**Solution:**

**Question 6:**

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that

∠CBD +∠CDB = ½ ∠BAD.**Solution:**

**Question 7:**

O is the circumcentre of the ΔABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.**Thinking Process**

Firstly, prove that ΔBOD and ΔCOD are congruent by SSS rule. Further, use the theorem that in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle and prove the required result.**Solution:**

Given In a ΔABC a circle is circumscribed having centre O.

**Question 8:**

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.**Solution:**

**Question 9:**

Two chords AB and AC of a circle subtends angles equal to 90° and 150°, respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.**Solution:**

**Question 10:**

If BM and CN are the perpendiculars drawn on the sides AC and AB of the ΔABC, prove that the points B, C, M and N are concyclic.**Solution:**

Given In ΔABC, BM ⊥ AC and CN ⊥ AB.

To prove Points B, C, M and N are con-cyclic.

Construction Draw a circle passing through the points B, C, M and N.

Proof Suppose, we consider SC as a diameter of the circle. Also, we know that SC subtends a 90° to the circle.

So, the points M and N should be on a circle.

Hence, BCMN form a con-cyclic quadrilateral. Hence proved.

**Question 11:**

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic.**Solution:**

Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC.

To prove Quadrilateral BCDE is a cyclic quadrilateral.

Construction Draw a circle passes through the points B, C, D and E.

**Question 12:**

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal.**Thinking Process**

Firstly, prove that ΔACD is congruent to ΔBDC by SAS and then further prove the required result**Solution:**

Given Let ABCD be a cyclic quadrilateral and AD = BC.

Join AC and BD.

To prove AC = BD

Proof In ΔAOD and ΔBOC,

∠OAD = ∠OBC and ∠ODA = ∠OCB

[since, same segments subtends equal angle to the circle]

AB = BC [given]

ΔAOD = ΔBOC [by ASA congruence rule]

Adding is DOC on both sides, we get

ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC

=> ΔADC ≅ ΔBCD

AC = BD [by CPCT]

**Question 13:**

The circumcentre of the ΔABC is 0. Prove that ∠OBC + ∠BAC = 90°.**Solution:**

Given A circle is circumscribed on a ΔABC having centre O.

**Question 14:**

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.**Solution:**

Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e.,

AB = BO …**(i)**

Join OA, AC and BC.

Since, OA = OB= Radius of circle

OA = AS = BO

**Question 15:****In figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.****Solution:**

We have, ∠ADC = 130° and chord BC – chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral.

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

∠ADC + ∠OBC = 180°

=> 130°+ ∠OBC = 180°

=> ∠OBC = 180° -130° = 50°

In ΔBOC and ΔBOE,

BC = BE [given equal chord]

OC =OE [both are the radius of the circle]

and OB = OB [common side]

ΔBOC ≅ΔBOE

∠OBC = ∠OBE = 50° [by CPCT]

∠CBE = ∠CBO + ∠EBO = 50° + 50° = 100°

**Question 16:****In figure, ∠ACB = 40°. Find ∠OAB.****Solution:**

**Question 17:**

A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130°. Find ∠BAC.**Solution:**

**Question 18:**

Two circles with centres O and O’ intersect at two points A and B. A line PQ is drawn parallel to OO’ through A (or B) intersecting the circles at P and Q. Prove that PQ =2 OO’.**Solution:**

Given, draw two circles having centres O and O’ intersect at points A and 8.

**Question 19:****In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.****Solution:**

Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral.

∠ACD+ ∠AED = 180° …**(i)** [sum of opposite angles in a cyclic quadrilateral is 180°]

**Question 20:****In figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.****Solution:**

**Exercise 10.4: Long Answer Type Questions**

**Question 1:**

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.**Solution:**

**Question 2:**

If non-parallel sides of a trapezium are equal, prove that it is cyclic.**Solution:**

Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

**Question 3:**

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic.**Solution:**

**Question 4:**

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.**Solution:**

Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic.

Construction Join PQ

Proof ∠1 = ∠A [exterior angle property of cyclic quadrilateral]

But ∠A = ∠C [opposite angles of a parallelogram]

∴ ∠1 = ∠C ,..(i)

But ∠C+ ∠D = 180° [sum of cointerior angles on same side is 180°]

=> ∠1+ ∠D = 180° [**from Eq. (i)**]

Thus, the quadrilateral QCDP is cyclic.

So, the points P, Q, C and D are con-cyclic. Hence proved.

**Question 5:**

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle.**Solution:**

**Question 6:****If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle.****Solution:**

Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles.

To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle.

Construction Draw a diameter EF parallel to CD having centre M.

Proof Since, CD||EF

arc EC = arc PD …** (i)**

arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF …**(ii)**

**Question 7:**

If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, then prove that PA is angle bisector of ∠BPC.**Solution:**

Given ΔABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C.

To prove PA is an angle bisector of ∠BPC.

Construction Join PB and PC.

**Question 8:****In figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC = ½ (angle subtended by arc C x A at centre + angle subtended by arc DYB at the centre).****Solution:**

Given In a figure, two chords AB and CD intersecting each other at point E.

**Question 9:****If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.****Thinking Process**

Use the property of cyclic quardrilateral, the sum of opposite angles of cyclic quadrilateral is supplementary. Further, simplify it to prove the required result.**Solution:**

Given, ABCD is a cyclic quadrilateral.

DP and QB are the bisectors of ∠D and ∠B, respectively.

To prove PQ is the diameter of a circle.

Construction Join QD and QC.

**Question 10:**

A circle has radius √2 cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45°.**Solution:**

**Question 11:**

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.**Solution:**

**Question 12:****AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q ^{2} = p^{2}+ 3r^{2}.**

**Thinking Process**

Firstly, use the Pythagoras theorem in ΔOAM and ΔOAN and further adjust them to prove the required result.

**Solution:**

**Question 13:****In figure, O is the centre of the circle ∠BCO = 30°. Find x and y.****Solution:**

**Question 14:****In figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.****Solution:**

**All Chapter NCERT Exemplar Problems Solutions For Class 9 maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 9**

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