Circle — Exercise 16(A)

A chord of length 6 cm is drawn in a circle of diameter 10 cm, its distance from the center of the circle is :

1. 6 cm

2. 8 cm

3. 4 cm

4. 10 cm

Answer

Let AB be the chord of the circle with center O.

Given,

Diameter = 10 cm

Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{10}{2}$ = 5 cm.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AD = $\dfrac{AB}{2} = \dfrac{6}{2}$ = 3 cm.

In right angled triangle OAD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OD² + AD²

⇒ 5² = OD² + 3²

⇒ OD² = 5² - 3²

⇒ OD² = 25 - 9

⇒ OD² = 16

⇒ OD = $\sqrt{16}$ = 4 cm.

Hence, Option 3 is the correct option.

The given figure shows two concentric circles and AD is a chord. The relation between AB and CD is :

1. AB = CD

2. AB > CD

3. AB < CD

4. AB ≠ CD

Answer

Draw OE ⊥ AD.

We know that,

Perpendicular from center to chord, bisects the chord.

In smaller circle,

⇒ BE = EC ...........(1)

In larger circle,

⇒ AE = ED ............(2)

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

⇒ AE - BE = ED - EC

⇒ AB = CD.

Hence, Option 1 is the correct option.

In the given figure, chord AB is larger than chord CD. The relation between OM and ON is :

1. OM = ON

2. OM < ON

3. OM > ON

4. OM + ON = AB

Answer

Join OC and OB.

Given,

⇒ AB > CD

⇒ $\dfrac{AB}{2} \gt \dfrac{CD}{2}$

⇒ BM > CN

From figure,

OC = OB = radius = r.

In right angle triangle ONC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = ON² + CN²

⇒ r² = ON² + CN²

⇒ ON² = r² - CN²

⇒ ON = $\sqrt{r^2 - CN^2}$

In right angle triangle OMB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OM² + BM²

⇒ r² = OM² + BM²

⇒ OM² = r² - BM²

⇒ OM = $\sqrt{r^2 - BM^2}$

Since, BM > CN

∴ r² - BM² < r² - CN²

⇒ $\sqrt{r^2 - BM^2} \lt \sqrt{r^2 - CN^2}$

⇒ OM < ON.

Hence, Option 2 is the correct option.

The line joining the mid-points of two chords of a circle passes through its center, then the chords are :

1. not parallel to each other

2. equal to each other

3. parallel to each other

4. not equal to each other

Answer

The line from centre of circle to the midpoint of any chord is also the perpendicular bisector of the chord.

∴ The two chords are parallel as they are both perpendicular to the line through the centre.

Hence, Option 3 is the correct option.

In the given figure, O and O' are centers of two circles, AB // CD // OO', then which of the following is not true :

1. AB = 2 × OO'

2. CD = 2 × OO'

3. AB = CD

4. AB ≠ CD

Answer

Draw ON and O'M perpendicular to AB.

We know that,

Perpendicular from center to the chord bisects it.

∴ AN = NP and MP = MB

From figure,

⇒ AB = AN + PN + MP + MB

⇒ AB = PN + PN + MP + MP

⇒ AB = 2PN + 2MP

⇒ AB = 2(PN + MP)

⇒ AB = 2NM

⇒ AB = 2OO' ............(1)

Draw OE and O'F perpendicular to CD.

We know that,

Perpendicular from center to the chord bisects it.

∴ CE = EQ and QF = FD

From figure,

⇒ CD = CE + EQ + QF + FD

⇒ CD = EQ + EQ + QF + QF

⇒ CD = 2EQ + 2QF

⇒ CD = 2(EQ + QF)

⇒ CD = 2EF

⇒ CD = 2OO' ............(2)

From equation (1) and (2), we get :

⇒ AB = CD.

Thus, AB ≠ CD is not true.

Hence, Option 4 is the correct option.

A chord of length 8 cm is drawn at a distance of 3 cm from the center of a circle. Calculate the radius of the circle.

Answer

Let AB = 8 cm be the chord of the circle with center O.

We know that,

Perpendicular from center to chord, bisects the chord.

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

In right angled triangle OAC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OC² + AC²

⇒ OA² = 3² + 4²

⇒ OA² = 9 + 16

⇒ OA² = 25

⇒ OA = $\sqrt{25}$ = 5 cm.

Hence, radius of circle = 5 cm.

The radius of a circle is 17.0 cm and the length of perpendicular drawn from its center to a chord is 8.0 cm. Calculate the length of the chord.

Answer

Let AB be the chord of the circle with center O and OC be the perpendicular from center to the chord.

Given,

Radius (OA) = 17 cm

In right angled triangle OAC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OC² + AC²

⇒ 17² = 8² + AC²

⇒ AC² = 17² - 8²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$ = 15 cm.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AB = 2 × AC = 2 × 15 = 30 cm.

Hence, length of chord = 30 cm.

A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.

Answer

Let AB be the chord of the circle with center O at the distance of 5 cm from the center of the circle.

We know that,

Perpendicular from center to chord, bisects the chord.

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

In right angled triangle OAC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OC² + AC²

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

From figure,

OD = OA = 13 cm (Radius of same circle)

Let chord DE be at the distance of 12 cm from the center.

In right angled triangle OFD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OD² = OF² + FD²

⇒ 13² = 12² + FD²

⇒ FD² = 13² - 12²

⇒ FD² = 169 - 144

⇒ FD² = 25

⇒ FD = $\sqrt{25}$ = 5 cm.

⇒ DE = 2 × FD = 2 × 5 = 10 cm.

Hence, length of chord at a distance of 12 cm from the centre equals to 10 cm.

In the following figure, AD is a straight line. OP ⊥ AD and O is the centre of both the circles. If OA = 34 cm, OB = 20 cm and OP = 16 cm; find the length of AB.

Answer

From figure,

In right-angled triangle OBP,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OP² + BP²

⇒ 20² = 16² + BP²

⇒ 400 = 256 + BP²

⇒ BP² = 400 - 256

⇒ BP² = 144

⇒ BP = $\sqrt{144}$ = 12 cm.

In right-angled triangle AOP,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OP² + AP²

⇒ 34² = 16² + AP²

⇒ 1156 = 256 + AP²

⇒ AP² = 1156 - 256

⇒ AP² = 900

⇒ AP = $\sqrt{900}$ = 30 cm.

From figure,

AB = AP - BP = 30 - 12 = 18 cm.

Hence, AB = 18 cm.

In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Find the distance between the chords, if both the chords are :

(i) on the opposite sides of the center,

(ii) on the same side of the center.

Answer

(i) 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 = $\dfrac{AB}{2} = \dfrac{16}{2}$ = 8 cm, CE = $\dfrac{CD}{2} = \dfrac{30}{2}$ = 15 cm.

From figure,

OA = OC = radius = 17 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 17² = OE² + 15²

⇒ 289 = OE² + 225

⇒ OE² = 289 - 225

⇒ OE² = 64

⇒ OE = $\sqrt{64}$ = 8 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 17² = OF² + 8²

⇒ 289 = OF² + 64

⇒ OF² = 289 - 64

⇒ OF² = 225

⇒ OF = $\sqrt{225}$ = 15 cm.

From figure,

⇒ EF = OE + OF = 8 + 15 = 23 cm.

Hence, distance between the chords = 23 cm.

(ii) 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 = $\dfrac{AB}{2} = \dfrac{16}{2}$ = 8 cm, CE = $\dfrac{CD}{2} = \dfrac{30}{2}$ = 15 cm.

From figure,

OA = OC = radius = 17 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 17² = OE² + 15²

⇒ 289 = OE² + 225

⇒ OE² = 289 - 225

⇒ OE² = 64

⇒ OE = $\sqrt{64}$ = 8 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 17² = OF² + 8²

⇒ 289 = OF² + 64

⇒ OF² = 289 - 64

⇒ OF² = 225

⇒ OF = $\sqrt{225}$ = 15 cm.

From figure,

⇒ EF = OF - OE = 15 - 8 = 7 cm.

Hence, distance between the chords = 7 cm.

Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0 cm and the distance between the two chords is 21.0 cm; find the length of the other chord.

Answer

Let AB and CD be the two parallel chords.

Given,

Length of one chord is 24.0 cm. Let AB = 24 cm.

Draw OE ⊥ CD and OF ⊥ AB.

We know that,

Perpendicular from the center to the chord, bisects it.

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

From figure,

OA = OC = radius = $\dfrac{\text{Diameter}}{2} = \dfrac{30}{2}$ = 15 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 15² = OF² + 12²

⇒ 225 = OF² + 144

⇒ OF² = 225 - 144

⇒ OF² = 81

⇒ OF = $\sqrt{81}$ = 9 cm.

Given,

Distance between two chords = 21 cm

∴ FE = 21 cm

⇒ OE = FE - OF = 21 - 9 = 12 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 15² = 12² + CE²

⇒ 225 = 144 + CE²

⇒ CE² = 225 - 144

⇒ CE² = 81

⇒ CE = $\sqrt{81}$ = 9 cm.

As, perpendicular from center to the chord, bisects it.

∴ CD = 2 × CE = 2 × 9 = 18 cm.

Hence, length of other chord = 18 cm.

A chord CD of a circle, whose center is O, is bisected at P by a diameter AB.

Given OA = OB = 15 cm and OP = 9 cm. Calculate the lengths of :

(i) CD

(ii) AD

(iii) CB.

Answer

We know that,

A straight line drawn from the center of a circle to bisect a chord is at right angles to the chord.

∴ OP ⊥ CD.

Join OC, AD and BC.

Given,

OA = OB = 15 cm

∴ Radius of circle = 15 cm.

∴ OC = 15 cm.

(i) In right-angled triangle OCP,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OP² + CP²

⇒ 15² = 9² + CP²

⇒ 225 = 81 + CP²

⇒ CP² = 225 - 81

⇒ CP² = 144

⇒ CP = $\sqrt{144}$ = 12 cm.

Since, chord CD is bisected at point P.

∴ CD = 2 × CP = 2 × 12 = 24 cm.

Hence, CD = 24 cm.

(ii) Since, chord CD is bisected at point P.

∴ PD = CP = 12 cm.

From figure,

⇒ AP = OA + OP = 15 + 9 = 24 cm.

In right-angled triangle APD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AD² = AP² + PD²

⇒ AD² = 24² + 12²

⇒ AD² = 576 + 144

⇒ AD² = 720

⇒ AD = $\sqrt{720}$ = 26.83 cm.

Hence, AD = 26.83 cm.

(iii) From figure,

AB is the diameter of the circle.

∴ AB = 2 × OA = 2 × 15 = 30 cm.

⇒ PB = AB - AP = 30 - 24 = 6 cm.

In right-angled triangle CPB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ CB² = CP² + PB²

⇒ CB² = 12² + 6²

⇒ CB² = 144 + 36

⇒ CB² = 180

⇒ CB = $\sqrt{180}$ = 13.42 cm.

Hence, CB = 13.42 cm.

A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centers O and O'.

Prove that the chords AB and CD, which are intercepted by the two circles, are equal.

Answer

Draw OP ⊥ AB and O'Q ⊥ CD.

In △ OMP and △ O'MQ,

⇒ ∠OMP = ∠O'MQ (Vertically opposite angles are equal)

⇒ ∠OPM = ∠O'QM (Both equal to 90°)

⇒ OM = O'M (As, M is the mid-point of OO')

∴ △ OMP ≅ △ O'MQ (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ OP = O'Q.

We know that,

Two chords of a circle or equal circles which are equidistant from the center are equal.

∴ AB = CD.

Hence, proved that AB = CD.

M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O. Prove that :

(i) ∠BMN = ∠DNM

(ii) ∠AMN = ∠CNM.

Answer

Join OM, ON, OB and OD.

Given,

M and N are the mid-points of two equal chords AB and CD.

∴ BM = $\dfrac{1}{2}AB$ and DN = $\dfrac{1}{2}CD$

Since, AB and CD are equal chords.

∴ BM = DN ..........(1)

We know that,

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

∴ OM ⊥ AB and ON ⊥ CD.

∴ ∠OMB = ∠OND (Both equal to 90°) ........(2)

(i) In triangle OMN,

⇒ ON = OM (Equal chords of a circle are equidistant from the center.)

⇒ ∠OMN = ∠ONM (Angles opposite to equal sides are equal) ........(3)

Subtracting equation (2) from (3), we get :

⇒ ∠OMB - ∠OMN = ∠OND - ∠ONM

⇒ ∠BMN = ∠DNM.

Hence, proved that ∠BMN = ∠DNM.

(ii) From figure,

⇒ ∠OMA = ∠ONC (Both equal to 90°) ...............(4)

Adding equation (3) and (4), we get :

⇒ ∠OMA + ∠OMN = ∠ONC + ∠ONM

⇒ ∠AMN = ∠CNM.

Hence, proved that ∠AMN = ∠CNM.

Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle. 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 = x (let) ..............(1)

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.

Adding equation (1) to both the sides, 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 PB = CD.

In the following figure, OABC is a square. A circle is drawn with O as center which meets OC at P and OA at Q. Prove that :

(i) △ OPA ≅ △ OQC

(ii) △ BPC ≅ △ BQA

Answer

(i) In △ OPA and △ OQC,

⇒ OP = OQ (Radius of same circle)

⇒ ∠AOP = ∠COQ (Both equal to 90°)

⇒ OA = OC (Sides of square)

∴ △ OPA ≅ △ OQC (By S.A.S. axiom)

Hence, proved that △ OPA ≅ △ OQC.

(ii) Proved above,

⇒ OC = OA ...........(1)

⇒ OP = OQ ...........(2)

Subtracting equation (2) from (1),

⇒ OC - OP = OA - OQ

⇒ CP = QA

In △ BPC and △ BQA,

⇒ BC = BA (Sides of square)

⇒ ∠PCB = ∠QAB (Both equal to 90°)

⇒ CP = QA (Proved above)

∴ △ BPC ≅ △ BQA (By S.A.S. axiom)

Hence, proved that △ BPC ≅ △ BQA.

The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm and 34 cm, calculate the distance between their centers.

Answer

Let O be the center of the circle with diameter 50 cm and O' be the center of the circle with diameter 34 cm.

Let AB be the common chord.

Radius = $\dfrac{\text{Diameter}}{2}$

Radius of circle with center O = $\dfrac{50}{2}$ = 25 cm,

Radius of circle with center O' = $\dfrac{34}{2}$ = 17 cm.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AC = BC = $\dfrac{AB}{2} = \dfrac{30}{2}$ = 15 cm.

In right-angled triangle AOC,

By pythagoras theorem,

⇒ AO² = AC² + OC²

⇒ 25² = 15² + OC²

⇒ 625 = 225 + OC²

⇒ OC² = 625 - 225

⇒ OC² = 400

⇒ OC = $\sqrt{400}$ = 20 cm.

In right-angled triangle AO'C,

By pythagoras theorem,

⇒ AO'² = AC² + O'C²

⇒ 17² = 15² + O'C²

⇒ 289 = 225 + O'C²

⇒ O'C² = 289 - 225

⇒ O'C² = 64

⇒ O'C = $\sqrt{64}$ = 8 cm.

From figure,

OO' = OC + O'C = 20 + 8 = 28 cm.

Hence, distance between centers = 28 cm.

The line joining the mid-points of two chords of a circle passes through its center. Prove that the chords are parallel.

Answer

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

∴ OM ⊥ AB and ON ⊥ CD.

So,

⇒ ∠OMA = ∠OMB = 90° and ∠ONC = ∠OND = 90°

Since,

⇒ ∠OMA = ∠OND = 90° (Alternate angles are equal)

⇒ ∠OMB = ∠ONC = 90° (Alternate angles are equal)

Hence, proved that AB || CD.

In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles. Show that :

(i) AB = CD

(ii) AC = BD

Answer

We know that,

Perpendicular from center to chord, bisects the chord.

Since, ABCD is perpendicular to PQ.

∴ OA = OD ............(1)

Also,

⇒ OB = OC ..............(2)

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

⇒ OA - OB = OD - OC

⇒ AB = CD.

Hence, proved that AB = CD.

(ii) We know that,

⇒ AB = CD

Adding BC to both sides of above equation, we get :

⇒ AB + BC = CD + BC

⇒ AC = BD.

Hence, proved that AC = BD.