Circle — Test Yourself

In a circle with center at point O, chord AB is a side of a square and chord BC is a side of regular hexagon. Then angle AOC is equal to:

1. 120°

2. 150°

3. 90°

4. none of these

Answer

We know that,

The angle subtended by each side of an n-sided regular polygon at the center of circle = $\dfrac{360°}{n}$

Given, AB is the side of the square.

Angle subtended by each arm of the square at the center of the circle is $\dfrac{360°}{4}$ = 90°.

⇒ ∠AOB = 90°.

BC is the side of the hexagon.

Angle subtended by each arm of the hexagon at the center of the circle is $\dfrac{360°}{6}$ = 60°.

⇒ ∠BOC = 60°.

From figure,

∠AOC = ∠AOB + ∠BOC = 90° + 60° = 150°.

Hence, option 2 is the correct option.

AB (= 20 cm) is diameter of the given circle and AP (= 16 cm). The distance of chord AP from center O is:

1. 12 cm

2. 18 cm

3. 9 cm

4. 6 cm

Answer

Given:

Length of the chord AP = 16 cm.

Diameter of the circle AB = 20 cm.

Radius of the circle, r = $\dfrac{20}{2}$ = 10 cm.

Construction: Draw OC ⊥ AP, where O is the center of the circle.

Since, perpendicular drawn from the center of a circle to a chord bisects it.

∴ OC bisects AP

AC = $\dfrac{1}{2}$ x AP = $\dfrac{1}{2}$ x 16 = 8 cm.

In Δ OAC, ∠C = 90°

Using Pythagoras theorem,

∴ OA² = OC² + AC²

⇒ (10)² = OC² + (8)²

⇒ 100 = OC² + 64

⇒ OC² = 100 - 64

⇒ OC² = 36

⇒ OC = ​$\sqrt{36}$

⇒ OC = 6 cm.

So, the distance of the chord from the center of the circle is 6 cm.

Hence, option 4 is the correct option.

Given O is center of the circle with chord AB = 8 cm, OA = 5 cm and OD ⊥ AB. The length of CD is :

1. 3 cm

2. 5 cm

3. 2 cm

4. none of these

Answer

Given, the length of chord AB = 8 cm.

OD ⊥ AB.

Since, perpendicular drawn from the center of a circle to a chord bisects it.

∴ OC bisects AB.

⇒ AC = $\dfrac{AB}{2} = \dfrac{8}{2}$ = 4 cm

Radius of the circle, OA = 5 cm.

In Δ OAC, ∠C = 90°

Using Pythagoras theorem,

∴ OA² = OC² + AC²

⇒ 5² = OC² + 4²

⇒ 25 = OC² + 16

⇒ OC² = 25 - 16

⇒ OC² = 9

⇒ OC = $\sqrt{9}$

⇒ OC = 3 cm

Since, OD = OA (Radii of the circle)

From figure,

⇒ CD = OD - OC = 5 - 3 = 2 cm.

Hence, option 3 is the correct option.

AB and CD are the chords of a circle with centre O, ∠AOB = 60° and angles ∠COD = 45°; the ratio between the length of the chords AB and CD is

1. 3 : 4

2. 4 : 3

3. 7 : 4

4. 7 : 3

Answer

We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the length of the chords.

$\therefore \dfrac{\text{∠AOB}}{\text{∠COD}} = \dfrac{\text{AB}}{\text{CD}} \\[1em] \Rightarrow \dfrac{60°}{45°} = \dfrac{\text{AB}}{\text{CD}} \\[1em] \Rightarrow \dfrac{4}{3} = \dfrac{\text{AB}}{\text{CD}}$

So, ratio between the length of the chords AB and CD is 4 : 3.

Hence, option 2 is the correct option.

Statement 1: O and O' are centres of two equal circles and ABCD is a straight line.

Statement 2: If OP ⊥ AB, O'Q ⊥ CD and O'Q is greater than OP, then CD > AB.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

O and O' are centres of two equal circles means they are having same radius.

The points A, B, C and D are shown aligned horizontally forming a single, continuous line.

∴ Statement 1 is true.

According to fundamental property of a circle, in two same circles, the chord that is farther from the center is shorter, and the chord that is closer to the center is longer.

Given, OP ⊥ AB, O'Q ⊥ CD and O'Q > OP.

⇒ CD < AB

∴ Statement 2 is false.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Statement 1: In a circle with center O, chord AB : chord BC = 1 : 3. If angle AOC is 160° ⇒ angle BOC = 120°.

Statement 2: AB : BC = 1 : 3

⇒ ∠AOC = 3 x ∠AOB

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the length of the chords.

$\therefore \dfrac{\text{∠AOB}}{\text{∠BOC}} = \dfrac{\text{AB}}{\text{BC}}\\[1em] \Rightarrow \dfrac{\text{∠AOB}}{\text{∠BOC}} = \dfrac{1}{3}\\[1em] \Rightarrow 3 \times \text{∠AOB} = \text{∠BOC}.$

So, statement 2 is false.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC

⇒ 160° = ∠AOB + 3∠AOB

⇒ 160° = 4∠AOB

⇒ ∠AOB = $\dfrac{160°}{4}$

⇒ ∠AOB = 40° and, ∠BOC = 3 x 40° = 120°.

So, statement 1 is true.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Assertion (A): In the given figure, chord AB = 8 cm, diameter CD = 20 cm, then length of OP = 10 cm.

Reason (R): OP = $\sqrt{OA^2 - AP^2}$

and CP = OC + OP

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given:

Length of the chord AB = 8 cm.

Diameter of the circle CD = 20 cm.

As we know that the radius of a circle is exactly half its diameter.

Radius of the circle, r = $\dfrac{20}{2}$ = 10 cm.

Construction: Join OA.

OP ⊥ AB.

Since, perpendicular drawn from the center of a circle to a chord bisects it.

∴ OP bisects AB

⇒ AP = $\dfrac{1}{2}$ x AB = $\dfrac{1}{2}$ x 8 = 4 cm

⇒ OA = 10 cm

In Δ OAP, ∠P = 90°

Using Pythagoras theorem,

∴ OA² = OP² + AP²

⇒ OP² = OA² - AP²

$\Rightarrow OP = \sqrt{OA^2 - AP^2}\\[1em] \Rightarrow OP = \sqrt{10^2 - 4^2}\\[1em] \Rightarrow OP = \sqrt{100 - 16}\\[1em] \Rightarrow OP = \sqrt{84}\\[1em] \Rightarrow OP = 2\sqrt{21} \text{ cm}$

∴ Assertion (A) is false.

From figure, CP = CO + OP = 10 + 2 $\sqrt{21}$ cm

∴ Reason (R) is true.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

The figure, given below, shows a circle with center O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm, find the radius of the circle.

Answer

Join OD.

Let radius of circle be x cm.

From figure,

Radius = OB = OD = x cm

⇒ OE = OB - EB = (x - 4) cm.

We know that,

A straight line drawn from the center of a circle to bisect a chord, is perpendicular to the chord.

∴ OE ⊥ CD

In right-angled triangle OED,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OD² = OE² + ED²

⇒ x² = (x - 4)² + 8²

⇒ x² = x² + 4² - 2 × x × 4 + 64

⇒ x² = x² + 16 - 8x + 64

⇒ x² - x² + 8x = 80

⇒ 8x = 80

⇒ x = $\dfrac{80}{8}$ = 10 cm.

Hence, radius of circle = 10 cm.

In the given figure, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the :

(i) the radius of the circle

(ii) length of chord CD.

Answer

(i) Join OA.

We know that,

Perpendicular from center to chord, bisects the chord.

As, OM ⊥ AB

∴ AM = $\dfrac{AB}{2} = \dfrac{24}{2}$ = 12 cm.

In △ OAM,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OM² + AM²

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

Hence the radius of circle = 13 cm.

(ii) We know that,

Perpendicular from center of the circle to chord, bisects the chord.

As, ON ⊥ CD

∴ CN = $\dfrac{CD}{2}$

⇒ CD = 2CN .............(1)

Join OC.

In △ OCN,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = ON² + NC²

⇒ 13² = 12² + CN²

⇒ CN² = 13² - 12²

⇒ CN² = 169 - 144

⇒ CN² = 25

⇒ CN = $\sqrt{25}$ = 5 cm.

Substituting value of CN in equation (1), we get :

⇒ CD = 2 × 5 = 10 cm.

Hence, length of chord CD = 10 cm.

AB and CD are two equal chords of a circle with center O which intersect each other at right angle at point P. If OM ⊥ AB and ON ⊥ CD; show that OMPN is a square.

Answer

Since, OM ⊥ AB and ON ⊥ CD.

∴ ∠OMP = ∠ONP = 90°.

Given,

AB and CD intersect at right angle.

∴ ∠MPN = 90°.

We know that,

Sum of angles of quadrilateral equal to 360°.

∴ ∠OMP + ∠ONP + ∠MPN + ∠MON = 360°

⇒ 90° + 90° + 90° + ∠MON = 360°

⇒ 270° + ∠MON = 360°

⇒ ∠MON = 360° - 270° = 90°.

Given,

AB = CD

We know that,

Equal chords are equidistant from the center.

∴ OM = ON = x (let) ............(1)

Since,

⇒ AB ⊥ CD and ON ⊥ CD

∴ AB || ON

∴ MP || ON.

⇒ OM ⊥ AB and CD ⊥ AB

∴ OM || CD

∴ OM || PN.

Since, in quadrilateral OMPN opposite sides are parallel.

∴ OMPN is parallelogram.

We know that,

Opposite sides of parallelogram are equal.

∴ MP = ON = x and OM = PN = x

∴ MP = ON = OM = PN.

Since, all sides of quadrilateral OMPN are equal and each interior angle to 90°.

Hence, proved that OMPN is a square.

The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centers.

Answer

Let AB be the chord of the circle with center O. Draw OC ⊥ AB.

We know that,

Perpendicular from the center of circle to the chord, bisects the chord.

∴ AC = $\dfrac{AB}{2} = \dfrac{24}{2}$ = 12 cm.

In right-angled triangle OAC,

⇒ OA² = OC² + AC²

⇒ 13² = OC² + 12²

⇒ 169 = OC² + 144

⇒ OC² = 169 - 144

⇒ OC² = 25

⇒ OC = $\sqrt{25}$ = 5 cm.

Hence, the distance of the chord from the center = 5 cm.

Prove that equal chords of congruent circles subtend equal angles at their centers.

Answer

We know that,

Congruent circles have equal radius.

Let there be two congruent circles with center O and O' and radius equal to r units.

Let there be two equal chords AB and CD.

In △ AOB and △ CO'D,

⇒ OA = O'C (Both equal to r units)

⇒ OB = O'D (Both equal to r units)

⇒ AB = CD (Given)

⇒ △ AOB ≅ △ CO'D (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠AOB = ∠CO'D.

Hence, proved that equal chords of congruent circles subtend equal angles at their centers.

Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points ?

Answer

Let there be two circles of different radii with centers A and C.

There can be four cases with two circles of different radii :

(i) No point of intersection when circles are away from each other.

(ii) One point of intersection when circles touch each other.

(iii) Two points of intersection when circles intersect each other.

(iv) Infinite, when one circle is completely inside the other and centers are concurrent.

Hence, maximum number of common points are infinite.

Suppose you are given a circle. Describe a method by which you can find the center of this circle.

Answer

Steps of construction :

1. Draw two chords AB and CD in the given circle.

2. Draw XY, the perpendicular bisector of chord AB and MN, the perpendicular bisector of chord CD.

3. Mark point O, the intersection of perpendicular bisectors of both the chords.

Point O, is the center of the circle

Given two equal chords AB and CD of a circle, with center O, intersecting each other at point P. Prove that :

(i) AP = CP

(ii) BP = DP

Answer

Draw, OM ⊥ AB and ON ⊥ CD.

Join OP, OB and OD.

We know that,

Perpendicular to a chord, from the center of the circle, bisects the chord.

∴ OM and ON bisects AB and CD respectively.

Given,

Two chords are equal.

∴ AB = CD = x (let)

∴ MB = $\dfrac{1}{2}AB = \dfrac{1}{2}x$ and ND = $\dfrac{1}{2}CD = \dfrac{1}{2}x$,

∴ MB = ND ..............(1)

Let, MB = ND = x.

From figure,

OB = OD = y (Radius of same circle)

In right-angled triangle OMB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OM² + MB²

⇒ OM² = OB² - MB²

⇒ OM² = y² - x²

⇒ OM = $\sqrt{y^2 - x^2}$ ........(2)

In right-angled triangle OND,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OD² = ON² + ND²

⇒ ON² = OD² - ND²

⇒ ON² = y² - x²

⇒ ON = $\sqrt{y^2 - x^2}$ ........(3)

From equation (2) and (3), we get :

⇒ OM = ON

In △ OPM and △ OPN,

⇒ ∠OMP = ∠ONP (Both equal to 90°)

⇒ OP = OP (Common side)

⇒ OM = ON (Proved above)

∴ △ OPM ≅ △ OPN (By R.H.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ PM = PN ..........(2)

Subtracting equation (2) from (1), we get :

⇒ MB - PM = ND - PN

⇒ PB = PD ...........(4)

Given,

⇒ AB = CD .........(5)

Subtracting equation (4) from (5), we get :

⇒ AB - PB = CD - PD

⇒ AP = CP.

Hence, proved that AP = CP and BP = DP.

In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. Calculate the distance between the chords, if they are on :

(i) the same side of the center

(ii) the opposite side of the center

Answer

(i) Let AB and CD be chords on same side of the center of the circle.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{12}{2}$ = 6 cm, CE = $\dfrac{CD}{2} = \dfrac{16}{2}$ = 8 cm.

From figure,

OA = OC = radius = 10 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 10² = OE² + 8²

⇒ 100 = OE² + 64

⇒ OE² = 100 - 64

⇒ OE² = 36

⇒ OE = $\sqrt{36}$ = 6 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 10² = OF² + 6²

⇒ 100 = OF² + 36

⇒ OF² = 100 - 36

⇒ OF² = 64

⇒ OF = $\sqrt{64}$ = 8 cm.

From figure,

⇒ EF = OF - OE = 8 - 6 = 2 cm.

Hence, distance between the chords = 2 cm.

(ii) Let AB and CD be chords on opposite side of the center of the circle.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\dfrac{AB}{2} = \dfrac{12}{2}$ = 6 cm, CE = $\dfrac{CD}{2} = \dfrac{16}{2}$ = 8 cm.

From figure,

OA = OC = radius = 10 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 10² = OE² + 8²

⇒ 100 = OE² + 64

⇒ OE² = 100 - 64

⇒ OE² = 36

⇒ OE = $\sqrt{36}$ = 6 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 10² = OF² + 6²

⇒ 100 = OF² + 36

⇒ OF² = 100 - 36

⇒ OF² = 64

⇒ OF = $\sqrt{64}$ = 8 cm.

From figure,

⇒ EF = OE + OF = 6 + 8 = 14 cm.

Hence, distance between the chords = 14 cm.

In the given figure, O is the center of the circle with radius 20 cm and OD is perpendicular to AB.

If AB = 32 cm, find the length of CD.

Answer

Join OA.

We know that,

Perpendicular from centre to the chord, bisects the chord.

∴ AC = $\dfrac{AB}{2} = \dfrac{32}{2}$ = 16 cm.

In right-angled triangle OAC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OC² + AC²

⇒ 20² = OC² + 16²

⇒ 400 = OC² + 256

⇒ OC² = 400 - 256

⇒ OC² = 144

⇒ OC = $\sqrt{144}$ = 12 cm.

From figure,

CD = OD - OC = 20 - 12 = 8 cm.

Hence, CD = 8 cm.

In the given figure, AB and CD are two equal chords of a circle, with center O.

If P is the mid-point of chord AB. Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

Answer

We know that,

Equal chords are equidistant from the center.

∴ OP = OQ

In △ OPQ,

⇒ OP = OQ

⇒ ∠OQP = ∠OPQ = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OQP + ∠OPQ + ∠POQ = 180°

⇒ x + x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$ = 15°.

We know that,

A straight line drawn from the center of the circle to bisect the chord, which is not a diameter, is at right angles to the chord.

∴ ∠OPA = 90°

From figure,

⇒ ∠APQ = ∠OPA - ∠OPQ = 90° - x = 90° - 15° = 75°.

Hence, ∠APQ = 75°.

In the given figure, AOC is the diameter of the circle, with center O. If arc AXB is half of arc BYC, find ∠BOC.

Answer

We know that,

Ratio of the angles subtended by the arcs on the center is equal to the ratio of the arcs.

$\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AXB}{BYC} \\[1em] \Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{1}{2} \\[1em] \Rightarrow ∠AOB = \dfrac{∠BOC}{2}.$

From figure,

⇒ ∠AOB + ∠BOC = 180°

⇒ $\dfrac{∠BOC}{2}$ + ∠BOC = 180°

⇒ $\dfrac{2∠BOC + ∠BOC}{2}$ = 180°

⇒ $\dfrac{3∠BOC}{2}$ = 180°

⇒ ∠BOC = $\dfrac{2}{3} \times 180° = \dfrac{360°}{3}$ = 120°.

Hence, ∠BOC = 120°.

The circumference of a circle, with center O, is divided into three arcs APB, BQC and CRA such that :

$\dfrac{\text{arc APB}}{2} = \dfrac{\text{arc BQC}}{3} = \dfrac{\text{arc CRA}}{4}$

Find ∠BOC.

Answer

Given,

⇒ $\dfrac{\text{arc APB}}{2} = \dfrac{\text{arc BQC}}{3} = \dfrac{\text{arc CRA}}{4}$ = k (let)

⇒ arc APB = 2k, arc BQC = 3k and arc CRA = 4k.

We know that,

Ratio of the angles subtended by the arcs on the center is equal to the ratio of the arcs.

⇒ ∠AOB : ∠BOC : ∠COA = 2k : 3k : 4k

⇒ ∠AOB : ∠BOC : ∠COA = 2 : 3 : 4

⇒ ∠AOB = 2x°, ∠BOC = 3x°, ∠COA = 4x°.

From figure,

⇒ ∠AOB + ∠BOC + ∠COA = 360°

⇒ 2x° + 3x° + 4x° = 360°

⇒ 9x° = 360°

⇒ x° = $\dfrac{360°}{9}$ = 40°.

⇒ ∠AOB = 2x° = 2(40°) = 80°,

⇒ ∠BOC = 3x° = 3(40°) = 120°,

⇒ ∠COA = 4x° = 4(40°) = 160°.

Hence, ∠BOC = 120°.