Circle

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

Given,

Diameter = 10 cm

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

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AD = $\frac{AB}{2} = \frac{6}{2}$ = 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.

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.

Join OC and OB.

Given,

⇒ AB > CD

⇒ $\frac{AB}{2} \gt \frac{CD}{2}$

⇒ 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 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.

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.

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 = $\frac{AB}{2} = \frac{8}{2}$ = 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.

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.

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 = $\frac{AB}{2} = \frac{24}{2}$ = 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.

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.

(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 = $\frac{AB}{2} = \frac{16}{2}$ = 8 cm, CE = $\frac{CD}{2} = \frac{30}{2}$ = 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 = $\frac{AB}{2} = \frac{16}{2}$ = 8 cm, CE = $\frac{CD}{2} = \frac{30}{2}$ = 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.

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 = $\frac{AB}{2} = \frac{24}{2}$ = 12 cm.

From figure,

OA = OC = radius = $\frac{\text{Diameter}}{2} = \frac{30}{2}$ = 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.

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.

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.

Join OM, ON, OB and OD.

Given,

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

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

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.

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 = $\frac{1}{2}AB = \frac{1}{2}x$ and ND = $\frac{1}{2}CD = \frac{1}{2}x$,

∴ 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.

(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.

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 = $\frac{\text{Diameter}}{2}$

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

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

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AC = BC = $\frac{AB}{2} = \frac{30}{2}$ = 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.

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.

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.

We know that,

Equal arcs subtends equal chords.

Given,

⇒ arc APB = arc CQD

⇒ AB = CD.

Hence, Option 1 is the correct option.

We know that,

Greater the chord, greater is the angle subtended by it at the center.

Given,

⇒ ∠AOB < ∠COD

⇒ AB < CD

Hence, Option 2 is the correct option.

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

⇒ $\frac{∠AOB}{∠COD} = \frac{AB}{CD} = \frac{3}{2}$

⇒ ∠AOB : ∠COD = 3 : 2.

Hence, Option 2 is the correct option.

Since, ABC is an equilateral triangle.

∴ AB = BC = AC.

We know that,

Equal chords subtend equal angles at the center.

∴ ∠AOB = ∠BOC = ∠COA = x (let)

We know that,

Angles around a point add to 360 °.

∴ ∠AOB + ∠BOC + ∠COA = 360°

⇒ x + x + x = 360°

⇒ 3x = 360°

⇒ x = $\frac{360°}{3}$ = 120°

⇒ ∠AOB = 120°.

Hence, Option 4 is the correct option.

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

$$\begin{aligned}\Rightarrow \frac{∠AOB}{∠COD} = \frac{AB}{CD} \\[1em] \Rightarrow \frac{∠AOB}{60°} = \frac{5}{3} \\[1em] \Rightarrow ∠AOB = \frac{5}{3} \times 60° = \frac{300°}{3} = 100°.\end{aligned}$$

Hence, Option 3 is the correct option.

Join OA and OD.

(i) We know that,

Diagonals of a square bisect each other at 90°.

∴ ∠BOC = 90°.

Hence, ∠BOC = 90°.

(ii) From figure,

OB and OC are the radius of the circle.

In △ OBC,

⇒ OB = OC

⇒ ∠OCB = ∠OBC = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ x + x + 90° = 180°

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

⇒ ∠OCB = 45°.

Hence, ∠OCB = 45°.

(iii) We know that,

Diagonals of a square bisect each other at 90°.

∴ ∠COD = 90°.

Hence, ∠COD = 90°.

(iv) From figure,

⇒ ∠BOD = ∠BOC + ∠COD = 90° + 90° = 180°.

Hence, ∠BOD = 180° and BD is the diameter of the circle.

We know that,

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

(i) Given,

AB is the side of the pentagon.

Angle subtended by each arm of the pentagon at the center of the circle is $\frac{360°}{5}$ = 72°.

Hence, ∠AOB = 72°.

(ii) Given,

BC is the side of the hexagon.

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

Hence, ∠BOC = 60°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.

Hence, ∠AOC = 132°.

(iv) In △ OAB,

⇒ OA = OB

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 72° = 180°

⇒ 2x + 72° = 180°

⇒ 2x = 180° - 72°

⇒ 2x = 108°

⇒ x = $\frac{108°}{2}$

⇒ x = 54°

⇒ ∠OBA = 54°.

Hence, ∠OBA = 54°.

(v) In △ OBC,

⇒ OC = OB

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 60° = 180°

⇒ 2y + 60° = 180°

⇒ 2y = 180° - 60°

⇒ 2y = 120°

⇒ y = $\frac{120°}{2}$

⇒ y = 60°

⇒ ∠OBC = 60°.

Hence, ∠OBC = 60°.

(vi) From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 54° + 60° = 114°.

Hence, ∠ABC = 114°.

(i) We know that,

If two arcs are of a circle are equal, they subtend equal angles at the center.

∴ ∠BOC = ∠AOB = 48°.

Hence, ∠BOC = 48°.

(ii) Join BC.

In △ OBC,

⇒ OC = OB (Radius of same circle)

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 48° = 180°

⇒ 2y + 48° = 180°

⇒ 2y = 180° - 48°

⇒ 2y = 132°

⇒ y = $\frac{132°}{2}$

⇒ y = 66°

⇒ ∠OBC = 66°.

Hence, ∠OBC = 66°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°.

Hence, ∠AOC = 96°.

(iv) Join AC.

In △ AOC,

⇒ OC = OA (Radius of same circle)

⇒ ∠OAC = ∠OCA = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\frac{84°}{2}$

⇒ x = 42°

⇒ ∠OAC = 42°.

Hence, ∠OAC = 42°.

(i) We know that,

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

$$\begin{aligned}\Rightarrow \frac{∠AOB}{∠BOC} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{96°}{∠BOC} = \frac{3}{2} \\[1em] \Rightarrow ∠BOC = \frac{2}{3} \times 96° = 2 \times 32° = 64°.\end{aligned}$$

Hence, ∠BOC = 64°.

(ii) In △ AOB,

⇒ OA = OB (Radius of same circle)

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\frac{84°}{2}$

⇒ x = 42°

⇒ ∠OBA = 42°.

In △ BOC,

⇒ OB = OC (Radius of same circle)

⇒ ∠OCB = ∠OBC = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ y + y + 64° = 180°

⇒ 2y + 64° = 180°

⇒ 2y = 180° - 64°

⇒ 2y = 116°

⇒ y = $\frac{116°}{2}$

⇒ y = 58°

⇒ ∠OBC = 58°.

From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 42° + 58° = 100°.

Hence, ∠ABC = 100°.

(i) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠BOC = ∠AOB = 50°.

From figure,

⇒ ∠AOC = ∠BOC + ∠AOB = 50° + 50° = 100°.

Hence, ∠AOC = 100°.

(ii) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠DOC = ∠AOB = 50°.

From figure,

⇒ ∠AOD = ∠AOC + ∠DOC = 100° + 50° = 150°.

Hence, ∠AOD = 150°.

(iii) From figure,

⇒ ∠BOD = ∠BOC + ∠DOC = 50° + 50° = 100°.

Hence, ∠BOD = 100°.

(iv) Join AC.

In △ AOC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OAC = ∠OCA = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ y + y + 100° = 180°

⇒ 2y + 100° = 180°

⇒ 2y = 180° - 100°

⇒ 2y = 80°

⇒ y = $\frac{80°}{2}$

⇒ y = 40°

⇒ ∠OAC = 40°.

Hence, ∠OAC = 40°.

(v) Join AD.

In △ OAD,

⇒ OA = OD (Radius of same circle)

⇒ ∠ODA = ∠OAD = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAD + ∠ODA + ∠AOD = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\frac{30°}{2}$

⇒ x = 15°

⇒ ∠ODA = 15°.

Hence, ∠OAD = 15°.

We know that,

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

(i) Given,

AB is the side of the hexagon.

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

Hence, ∠AOB = 60°.

(ii) Given,

AC is the side of a regular eight sided polygon.

Angle subtended by each arm of the regular eight sided polygon at the center of the circle is $\frac{360°}{8}$ = 45°.

Hence, ∠AOC = 45°.

(iii) From figure,

⇒ ∠BOC = ∠AOB + ∠AOC = 60° + 45° = 105°.

Hence, ∠BOC = 105°.

(iv) In △ OBC,

⇒ OB = OC (Radius of same circle)

⇒ ∠OBC = ∠OCB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + 105° = 180°

⇒ 2x + 105° = 180°

⇒ 2x = 180° - 105°

⇒ 2x = 75°

⇒ x = $\frac{75°}{2}$

⇒ x = 37.5°

⇒ ∠OBC = 37.5°.

Hence, ∠OBC = 37.5°=37°30'.

(i) We know that,

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

$$\begin{aligned}\Rightarrow \frac{∠AOB}{∠BOC} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{100°}{∠BOC} = \frac{2}{1} \\[1em] \Rightarrow ∠BOC = \frac{1}{2} \times 100° = 50°.\end{aligned}$$

Hence, ∠BOC = 50°.

(ii) Join AC.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.

In △ OAC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OCA = ∠OAC = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\frac{30°}{2}$

⇒ x = 15°

⇒ ∠OAC = 15°.

Hence, ∠OAC = 15°.

We know that,

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

Given, AB is the side of the square.

Angle subtended by each arm of the square at the center of the circle is $\frac{360°}{4}$ = 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 $\frac{360°}{6}$ = 60°.

⇒ ∠BOC = 60°.

From figure,

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

Hence, option 2 is the correct option.

Given:

Length of the chord AP = 16 cm.

Diameter of the circle AB = 20 cm.

Radius of the circle, r = $\frac{20}{2}$ = 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 = $\frac{1}{2}$ x AP = $\frac{1}{2}$ 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, 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 = $\frac{AB}{2} = \frac{8}{2}$ = 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.

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.

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

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

Hence, option 2 is the correct option.

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.

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.

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

So, statement 2 is false.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC

⇒ 160° = ∠AOB + 3∠AOB

⇒ 160° = 4∠AOB

⇒ ∠AOB = $\frac{160°}{4}$

⇒ ∠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.

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 = $\frac{20}{2}$ = 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 = $\frac{1}{2}$ x AB = $\frac{1}{2}$ x 8 = 4 cm

⇒ OA = 10 cm

In Δ OAP, ∠P = 90°

Using Pythagoras theorem,

∴ OA² = OP² + AP²

⇒ OP² = OA² - AP²

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

∴ 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.

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 = $\frac{80}{8}$ = 10 cm.

Hence, radius of circle = 10 cm.

(i) Join OA.

We know that,

Perpendicular from center to chord, bisects the chord.

As, OM ⊥ AB

∴ AM = $\frac{AB}{2} = \frac{24}{2}$ = 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 = $\frac{CD}{2}$

⇒ 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.

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.

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 = $\frac{AB}{2} = \frac{24}{2}$ = 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.

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.

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.

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

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 = $\frac{1}{2}AB = \frac{1}{2}x$ and ND = $\frac{1}{2}CD = \frac{1}{2}x$,

∴ 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.

(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 = $\frac{AB}{2} = \frac{12}{2}$ = 6 cm, CE = $\frac{CD}{2} = \frac{16}{2}$ = 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 = $\frac{AB}{2} = \frac{12}{2}$ = 6 cm, CE = $\frac{CD}{2} = \frac{16}{2}$ = 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.

Join OA.

We know that,

Perpendicular from centre to the chord, bisects the chord.

∴ AC = $\frac{AB}{2} = \frac{32}{2}$ = 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.

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 = $\frac{30°}{2}$ = 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°.

We know that,

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

$$\begin{aligned}\Rightarrow \frac{∠AOB}{∠BOC} = \frac{AXB}{BYC} \\[1em] \Rightarrow \frac{∠AOB}{∠BOC} = \frac{1}{2} \\[1em] \Rightarrow ∠AOB = \frac{∠BOC}{2}.\end{aligned}$$

From figure,

⇒ ∠AOB + ∠BOC = 180°

⇒ $\frac{∠BOC}{2}$ + ∠BOC = 180°

⇒ $\frac{2∠BOC + ∠BOC}{2}$ = 180°

⇒ $\frac{3∠BOC}{2}$ = 180°

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

Hence, ∠BOC = 120°.

Given,

⇒ $\frac{\text{arc APB}}{2} = \frac{\text{arc BQC}}{3} = \frac{\text{arc CRA}}{4}$ = 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° = $\frac{360°}{9}$ = 40°.

⇒ ∠AOB = 2x° = 2(40°) = 80°,

⇒ ∠BOC = 3x° = 3(40°) = 120°,

⇒ ∠COA = 4x° = 4(40°) = 160°.

Hence, ∠BOC = 120°.