Circles Questions

Circles Questions are given here to help the students understand all basic concepts related to circles, and improve their problem-solving skills on applications of different theorems on circles. In this article, you will find questions on circles, along with solutions for each of these questions. Students can also practise the additional questions provided at the end of the article.

What is a circle?

In geometry, a circle is the collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane. The fixed point is called the centre, and the fixed distance is called the radius of the circle.

Learn more about circles here.

Circles Questions and Answers

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

Solution:

Given that arcs AXB and CYD of a circle are congruent,

i.e. arc AXB ≅ arc CYD.

We know that if two arcs of a circle are congruent, their corresponding chords are also equal.

i.e. chord AB = chord CD

Thus, AB/CD = 1

AB/CD = 1/1

AB : CD = 1 : 1

2. In the figure, AOC is the diameter of the circle and arc AXB = (1/2)arc BYC. Find ∠BOC.

Solution:

Given,

arc AXB = (1/2) arc BYC

∠AOB = (1/2) ∠BOC

Also, ∠AOB + ∠BOC = 180º

Therefore, (1/2) ∠BOC + ∠BOC = 180º {linear pair since AOC is the diameter}

(3/2) ∠BOC 180º

∠BOC = (2/3) × 180º = 120º

3. A circular park of a radius of 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at an 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.

Solution:

Let A, B, and C be the positions of Ankur, Syed and David, respectively.

As they are sitting at equal distances, the triangle is equilateral

AD ⊥ BC is drawn.

AD is the median of ΔABC, and it passes through the centre O.

O is the centroid of the ΔABC. OA is the radius of the triangle.

OA = 2/3 AD

Let a m be the side of the triangle.

So, BD = a/2 m

In ΔABD,

By Pythagoras theorem,

AB² = BD² + AD²

⇒ AD² = AB² – BD²

⇒ AD² = a² – (a/2)²

⇒ AD² = 3a²/4

⇒ AD = √3a/2

OA = 2/3 AD

20 m = (⅔) × (√3a/2)

⇒ a = 20√3 m

Therefore, the length of the string of the toy is 20√3 m.

4. In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution:

Given,

ABCD is a cyclic quadrilateral in which AC and BD are its diagonals.

∠DBC = 55° and ∠BAC = 45°

∠CAD = ∠DBC = 55° (Angles in the same segment)

Thus, ∠DAB = ∠CAD + ∠BAC

= 55° + 45°

= 100°

We know that the opposite angles of a cyclic quadrilateral are supplementary.

∠DAB + ∠BCD = 180°

∠BCD = 180° – 100° = 80°

5. The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:

Let O be the centre of the circle and AB is the tangent.

OB⊥ AB then ∆OAB is a right-angled triangle

OA² = AB² + OB² (by Pythagoras Theorem)

5² = 4² + OB²

OB² = 5² – 4²

= 25 – 16

OB² = 9

OB = √9 = 3

Hence, the radius of the circle = 3 cm.

6. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Solution:

Let C₁ and C₂ be the two circles having the same centre O.

AC is the chord that touches the C₁ at point D.

Join OD.

Also, OD is perpendicular to AC

∴ AD = DC = 4 cm [ Perpendicular line OD bisects the chord]

In right ΔAOD,

OA² = AD² + DO² {by Pythagoras theorem}

⇒ DO² = 5² − 4²

= 25 − 16

= 9

⇒ DO = 3

Therefore, the radius of the inner circle = 3 cm.

7. AB is a diameter, and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Solution:

Given:

O is the centre of a circle, AB is the diameter, and AC is the chord AC such that ∠BAC=30°.

A tangent from C meets AB at D on producing.

Join BC and OC.

∠BCD = ∠BAC = 30° (Angles in the alternate segment)

Arc BC subtends ∠DOC at the centre of the circle and ∠BAC at the remaining part of the circle.

∴∠BOC. =2∠BAC = 2 × 30° = 60°

In triangle OCD,

∠BOC or ∠DOC = 60° (Proved above)

∠OCD = 90° (∵ OC ⊥ CD)

∴∠DOC +∠ODC = 90°

⇒ 60° + ∠ODC = 90°

∴ ∠ODC = 90° − 60° = 30∘

In triangle BCD,

∠ODC or ∠BDC =∠BCD = 30°

∴BC = BD

Hence proved.

8. If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Solution:

Given:

In a circle, ΔABC is inscribed in which AB = AC = 6 cm and the radius of the circle is 9 cm

To Find: Area of triangle ABC

Construction: Join OB and OC

Proof:

In △AOB and △AOC,

OB = OC [Radii of same circle]

AO = AO [Common]

AB = AC = 6 cm [Given]

△AOB ≅ △AOC [By SSS congruence criterion]

∠OAB = ∠OBC [Corresponding parts of congruent triangles are equal ]

⇒ ∠MAB = ∠MBC

In △ABM and △AMC,

AB = AC = 6 cm [Given]

AM = AM [Common]

∠MAB = ∠MBC [Proved above]

△ABM ≅ △AMC [By SAS congruence criterion]

∠AMB = ∠AMC [Corresponding parts of congruent triangles are equal ]

Now,

∠AMB + ∠ AMC = 180°

∠AMB + ∠AMB = 180°

2 ∠AMB = 180

∠AMB = 90°

We know that a perpendicular from the centre of the circle bisects the chord.

So, OA is the perpendicular bisector of BC.

Let OM = x

AM = OA – OM = 9 – x [ As OA = radius = 9 cm]

In right angled ΔAMC,

(AM)² + (MC)² = (AC)² {By Pythagoras theorem}

(9 – x)² + (MC)² = (6)²

81 + x²– 18x + (MC)² = 36

(MC)² = 18x – x² – 45….(1)

In right △OMC ,

(MC)² + (OM)² = (OC)² {By Pythagoras Theorem}

18x – x² – 45 + (x)² = (9)²

18x – 45 = 81

18x = 126

x = 7

AM = 9 – x = 9 – 7 = 2 cm

In right △AMC,

(AM)² + (MC)² = (AC)² {By Pythagoras Theorem}

(2)² + (MC)² = (6)²

4 + (MC)² = 36

(MC)² = 32

MC = 4√2 cm

Also, MC = BM

BC = 2MC = 2(4√2) = 8√2 cm

Area of ΔABC = (1/2) × BC × AM

= (1/2) × 8√2 × 2

= 8√2 cm²

9. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution:

First, draw a circle and connect two points A and B such that AB becomes the diameter of the circle.

Now, draw two tangents PQ and RS at points A and B, respectively.

Now, both radii i.e. AO and OB are perpendicular to the tangents.

So, OB is perpendicular to RS and OA perpendicular to PQ

So, ∠OAP = ∠OAQ = ∠OBR = ∠OBS = 90°

From the above figure, angles OBR and OAQ are alternate interior angles.

Also, ∠OBR = ∠OAQ and ∠OBS = ∠OAP {since they are also alternate interior angles}

So, we can say that the line PQ and the line RS will be parallel to each other.

Hence proved.

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

Solution:

Given

Let circle C₁ have centre O and circle C₂ have centre O’.

PQ is the common chord.

To prove:

OO’ is the perpendicular bisector of PQ i.e.

1. PR = RQ

2. ∠PRO = ∠PRO’ = ∠QRO = ∠QRO’ = 90°

Construction:

Join PO, PO’, QO, QO’

Proof

In △POO’ and △QOO’,

OP = OQ (Radius of circle C1)

O’P = O’Q (Radius of circle C2)

OO’ = OO’ (Common)

∴ △POO’ ≅ △QOO’ (by SSS Congruence rule)

∠POO’ = ∠QOO’ (CPCT) —-(1)

Also,

In △POR and △QOR,

OP = OQ (Radius of circle C1)

∠POR = ∠QOR ( From (1))

OR = OR (Common)

∴ △OPO’ ≅ △OQO’ (by SAS Congruence Rule)

PR = QR (CPCT)

& ∠PRO = ∠QRO (CPCT) —-(2)

Since PQ is a line

∠PRO + ∠QRO = 180° (linear Pair)

∠PRO + ∠PRO= 180° ( From (2))

2∠PRO = 180°

∠PRO = 180° / 2

∠PRO = 90°

Therefore,

∠QRO = ∠PRO = 90°

Also,

∠PRO’ = ∠QRO = 90° (vertically opposite angles)

∠QRO’ = ∠PRO = 90° (vertically opposite angles)

Since, ∠PRO = ∠PRO’ = ∠QRO = ∠QRO’ = 90°

∴ OO’ is the perpendicular bisector of PQ.

Hence proved.

Video Lessons on Circles

Practice Questions on Circles

1. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
2. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
3. 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′.
4. A is a point at a distance of 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.
5. 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.