NCERT Exemplar Class 9 Maths Solutions Circles

Chapter 10 Circles solved by expert teachers as per NCERT (CBSE) Book guidelines. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations.

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
AO² = AC² + OC²   
(5)² = (4)² + OC²
25 = 16 + OC²
OC² = 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) 10 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 AC²+ BC² = AB².
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]
AC² + BC² = AB²

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

1. Firstly, inscribe a ΔABC in a circle, then draw the perpendicular bisecters of any two sides of a triangle.
2. 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 4q² = p²+ 3r².
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: