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Correct option (b) radius of director circle of S ≡  0 is √2

Explanation:

(x² + y² – 4y + 3) + λx = 0

∴ x = 0 and y² – 4y + 3 = 0

(y – 3)(y – 1) = 0

y = 3 , 1

∴ (0,3) (0,1) are common points. 

x² + y² + 2gx + 2fy + c = 0

passing through (0,3) & (0,1)

9 + 6f + c = 0 ...........(1)

1 + 2f  +  c = 0  ......... (2)

(1) – (2) we get

8 +  4f =  0

f = –2 and c = 3

x² + y² + 2gx – 4y + 3 = 0

radius = √(g² + 4 - 3) = √(g² + 1)

for minimum area radius must be minimum

Since g² + 1 is positive so g must be zero

∴ radius = 1

Area = πr² =  πsq.u.

Radius of director circle is √2 times the radius of the given circle. 

∴ Radius of director circle is √2

Correct option is: (D) 10 cm

\(\because\) Tangents drawn from a fixed outer point to circle are of equal length.

\(\therefore\) AP = AR, (Tangents drawn from outer point A)

 BP = BQ, (Tangents drawn from outer point B)

and CQ = CR, (Tangents drawn from outer point C)

Now, BC = BQ + QC

= BP + CR (\(\because\) BQ = BP and QC = CR)

= 3 + AC - AR (\(\because\) BP = 3cm (Given) & CR = AC - AR)

= 3 + 11 - PA (\(\because\) AC = 11 cm (given) & AR = AP = PA)

= 14 - 4 (\(\because\) PA = 4 cm (given))

= 10 cm.

Correct option is: (D) 10 cm

Correct option is: (A) 45°

From figure ∠ACB = 3x – 65° and- ∠AOB = 2x° 

From theorem, chord \(\overline {AB}\) making angle at centre ∠AOB = 2∠ACB 

∴ 2x = 2(3x – 65°) 

2x = 6x – 130° 

6x – 2x = 130° 

4x = 130°

x = 130°/4 = 32.5°

Correct option is (D) 2 ∠ACB

Angle subtended by an arc at the centre of the circle is double than the angle subtended by that arc at any point on the circumference.

\(\because\) \(\angle AOB\) = Angle subtended by arc AB at centre O

\(\angle ACB\) = Angle subtended by arc AB at point C of circumference

\(\therefore\) \(\angle AOB\) = \(2\,\angle ACB\)

Correct option is  D) 2 ∠ACB

Correct option is (B) Diameter

Diameter of a circle divides it into two semicircles.

Correct option is  B) Diameter

Correct option is (A) 1

There is only one circle can be drawn through three non-collinear points.

Correct option is  A) 1

(1) Exterior 

(2) on the circle 

(3) interior

Circles having the same centre are called concentric circles.

Correct option is (D) 3 cm

\(\angle AOB\) = Angle subtend by chord AB at the centre of the circle

\(\angle COD\) = Angle subtend by chord CD at the centre of the circle

Since, if two chord subtend equal angles at the centre of the circle then they are equal in length.

\(\because\) \(\angle AOB\) = \(\angle COD\)

\(\therefore\) AB = CD

\(\Rightarrow\) CD = AB = 3 cm  \((\because\) AB = 3 cm (given))

Correct option is  D) 3 cm

 Correct option is (B) 40°

\(\because\) arc AB \(\cong\) arc CD

Since, congruent (equal) arcs subtends equal angle the centre of the circle.

Also, \(\angle AOB\) = Angle subtended by arc AB at centre O

\(\angle COD\) = Angle subtended by arc CD at centre O

\(\therefore\) \(\angle AOB\) = \(\angle COD\)    \((\because arc\,AB\cong arc\,CD)\)

\(\Rightarrow\) \(\angle COD\) = \(\angle AOB\) \(=40^\circ\)   \((\because\angle AOB=40^\circ(given))\)

Correct option is  B) 40°

Correct answer is (b) 45°

We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees

∴ ∠AOD + ∠BOC = 180°

\(\Rightarrow\) ∠BOC = 180° - 135° = 45°

‘O’ is the centre of the circle. 

PQ = RS [given, equal chords] 

∴∠POQ = ∠ROS [ ∵ equal chords make equal angles at the centre] 

∴ In ΔROS ∠ORS + ∠OSR + ∠ROS = 180°

[angle sum property] 

∴ 48° + 48° + ∠ROS = 180° 

[ ∵ OR = OS(radii); ΔORS is isosceles] 

∴ ∠ROS = 180° – 96° = 84° 

Also ∠POQ = ∠ROS = 84° 

∴ ∠OPQ = ∠OQP 

[∵ OP = OQ; radii]

= 1/2 [180°-84°] = 48°

i) True 

ii) True 

iii) True 

iv) False 

v) False 

vi)True 

vii) True

O’ is the centre of the circle. 

AB = CD (equal chords from the figure) 

∴ ∠AOB = ∠COD 

[ ∵ equal chords make equal angles at the centre] 

∴ ∠COD = 90° [ ∵ ∠AOB = 90° given]

Answer : (c) 135º 

In ∆ DEF, ∠ EDF = 180° – (60° + 75°) 

= 180° – 135° = 45° 

∠OAD = ∠OBD = 90° (Tangents DE and DF are perpendicular to radii OA and OB respectively at A and B) 

∴ In quad . DAOB, ∠AOB = 360° – (90° + 90° + 45°) 

= 360° – 225° = 135°.

From the figure x = y [ ∵ angles opp. to opp. to equal sides] 

But x + y + 30° = 180°

∴ x + y = 180° – 30° = 150°

⇒ x + y = 150°/2 = 75°

Answer : (c) 58º 

OA = OB ⇒ ∠OBA = ∠OAB = 32° (Isosceles ∆ property) 

∠AOB = 180° – (∠OBA + ∠OAB) = 180° – 64° = 116° 

⇒ ∠ACB = \(\frac{1}{2}\) × ∠AOB) = \(\frac{1}{2}\) × 116° = 58° 

Also, ∠BAS = x = ∠ACB = 58°   (Angle in alternate segment are equal)

Option : (D)

Let AB and CD are two diameters of circle 

∠AOD = ∠BOD = ∠BOC = ∠AOC = 90° 

AB and CD are diagonals of quadrilateral ABCD 

They intersect each other at right angles 

And,

AB = BC = CD = DA 

We know that, 

Sides of a square are equal and diagonals intersect at 90° 

Therefore, 

ABCD is a square

Correct option is  B) 68°

‘O’ is the centre of the circle. 

AB, CD are equal chords 

⇒ They subtend equal angles at the centre. 

∴ ∠AOB =∠COD = 70° 

Now in ΔOCD 

∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides] 

∴ ∠OCD + ∠ODC + 70° = 180° 

= ∠OCD +∠ODC = 180° – 70° = 110° 

∴ ∠OCD + ∠ODC = 110° = 55°

Correct option is  B) 42°

Option : (D)

Given that, 

AB || CD 

(Chords on same side of centre) 

AO = CO (Radii) 

OL and OM perpendicular bisector of CD and AB respectively 

CL = LD = 6 cm 

AM = MB = 3 cm 

LM = 3 cm (Given) 

In COL, 

CO² = OL² + 6² ...(i) 

In AOM, 

AO² = AM² + OM² 

= 3² + (OL + LM)² 

= 9 + OL² + 9 + 6O L 

OL² = AO² - 18 - 6O L ... (ii) 

Using (ii) in (i), 

OL = 3 cm 

Putting OL in (i),

AO² = \(\sqrt{45}\)

 AO = 3\(\sqrt{5}\)

∠ADE = 95°(Given) 

Since, 

OA = OB, so 

∠OAB = ∠OBA 

∠OAB = 30° 

∠ADE + ∠ADC = 180° 

(Linear pair) 

95° + ∠ADC = 180° 

∠ADC = 85° 

We know that, 

∠ADC = 2∠ADC 

∠ADC = 2 x 85° 

∠ADC = 170° 

Since, 

AO = OC 

(Radii of circle) 

∠OAC = ∠OCA 

(Sides opposite to equal angle) ... (i) 

In triangle OAC, 

∠OAC + ∠OCA + ∠AOC = 180° 

∠OAC + ∠OAC + 170° = 180° 

[From (i)] 

2∠OAC = 10° 

∠OAC = 5° 

Thus, 

∠OAC = 5°

Option : (D)

Given that, 

AB || CD 

(Chords on opposite side of centre) 

DO = BO (Radii) 

OL and OM perpendicular bisector of CD and AB respectively 

LM = 23 cm 

AB = 16 cm 

In ΔOLB, 

OB² = OL2 + LB² 

OL² = 225 

OL = 15 cm 

OM = LM - OL 

= 8 cm 

In ΔOMD, 

OD² = OM² + MD² 

MD² = 225 

MD = 15 cm 

Now, 

CD = 2 MD 

= 30 cm

We know that,

∠ACB = \(\frac{∠APB }{2}\)

∠ACB = \(\frac{150 }{2}\)

∠ACB = 75°

Since, 

ACD is a straight line, so 

∠ACB + ∠BCD = 180° 

75° + ∠BCD = 180° 

∠BCD = 180° – 75° 

= 105° 

Now,

∠BCD =  \(\frac{1}{2}\)Reflex ∠BQD 

105° = (360° - ∠BQD) 

210o = 360° - ∠BQD 

∠BQD = 360° – 210° 

Therefore, 

∠BQD = 150°

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠APB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠APB

By substituting the values

∠ACB = 150/2

So we get

∠ACB = 75^o

We know that ACD is a straight line

It can be written as

∠ACB + ∠DCB = 180^o

By substituting the values

75^o + ∠DCB = 180^o

On further calculation

∠DCB = 180^o – 75^o

By subtraction

∠DCB = 105^o

We know that

∠DCB = ½ × reflex ∠BQD

By substituting the values

105^o = ½ × (360^o – x)

On further calculation

210^o = 36^o – x

By subtraction

x = 150^o

Therefore, the value of x is 150^o.

By degree measure theorem, 

∠AOB = 2∠ACB 

130° = 2∠ACB 

∠ACB = 65° 

Therefore, 

∠ACB + ∠BCD = 180° 

(Linear pair) 

65o + ∠BCD = 180° 

∠BCD = 115° 

By degree measure theorem, 

Reflex ∠BOD = 2∠BCD 

Reflex ∠BOD = 2 x 115° 

= 230° 

∠BOD + Reflex ∠BOD = 360° 

(Complete angle) 

230° + x = 360° 

x = 130°

Circumference (c) = 176 cm 

∴ 2πr = 176 

∴ 2 x 22/7 x r = 176 

∴ 44/7 x r = 176 

∴ r = 176 x 7/44 = 28 cm

∴ The radius of the circle is 28 cm.

Radius of the circular garden (r) = 56 m 

∴ Circumference of the circular garden (c) = 2πr 

= 2 x 22/7 x 56 

= 352 m 

∴ Length of the wire required to put 1- round fence = Circumference 

∴ Length of wire required to put a 4- round fence = 4 x Circumference 

= 4 x 352 

= 1408 m 

∴ Cost of wire per meter = Rs 40 

∴ Total cost = length of wire required x cost of the wire 

= 1408 x 40 

= Rs 56320 

∴ The cost to put a 4-round fence around the garden is Rs 56320.

It is given that AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E

We know that in a cyclic quadrilateral exterior angle is equal to the interior opposite angle.

It can be written as

Exterior ∠EDC = ∠A

Exterior ∠DCE = ∠B

We know that AB || CD

So we get

∠EDC = ∠B and ∠DCE = ∠A

We get

∠A = ∠B

Therefore, it is proved that △ AEB is isosceles.

i. Radius (r) = 7 cm

Diameter (d) = 2r 

= 2 x 7 = 14 cm 

Circumference (c) = πd 

= 22/7 x 14 

= 44 cm

ii. Diameter (d) = 28 cm 

Radius (r) = d/2 = 28/2 = 14 cm 

Circumference (c) = πd 

= 22/7 x 28 

= 88 cm

iii. Circumference (c) = 616 cm 

∴ πd = 616 

∴ 22/7 x d = 616 

∴ d = 616 x 7/22 

∴ d = 196 cm 

∴ Diameter (d) = 196 cm 

Radius (r) = d/2 = 196/2 = 98 cm

iv. Circumference (c) = 72.6 cm 

∴ πd = 72.6 

∴ 22/7 x d = 72.6 

∴  d = 72.6 x 7/22 

= 726/10 x 7/22 

= (33 x 7)/10

∴ d = 23.1 cm 

∴ Diameter (d) = 23.1 cm 

Radius (r) = d/2 = 23.1/2

= 11.55 cm

Measure of major arc = 360° – measure of corresponding minor arc 

∴ m (arc PYQ) = 360 – 110 

∴ m (arc PYQ) = 250° 

∴ The measure of the major arc PYQ is 250°

Diameter of the wheel of the bullock cart (d) = 1.4 m

Circumference of the wheel of the bullock cart (c) = πd

= 22/7 x 1.4 = 22/7 x 14/10 = 44/10 = 4.4m

Distance corvered in 1 ratation = Circumference of the wheel

= 4.4 m

Number of rotatiuons = (Distance covered)/ circumference 

= (1.1 km)/4.4m = (1.1 x 1000m)/4.4m  ...[1 km = 1000m]

= (11 x 1000)/44

= 250

The wheel of the bullock cart will compleate 250 rotations as the cart travels 1.1 km.

(C) 240°

Measure of major arc = 360° – measure of corresponding minor arc 

∴ m (arc AYB) = 360 – m (arc AXB)

∴ m (arc AYB) = 360 – 120 

∴ m (arc AYB) = 240°

Correct option is: Statement (2) is true.

i. OA, OB, OC, OF 

ii. EC, AD, AB, FC 

iii. AB, FC

 (d) two diameters

Two diameters cannot be parallel as they perpendicularly bisect each other.

Given : A circle with center O and radius = 5cm T is a point, OT = 13 cm. OT intersects the circle at E and AB is the tangent to the circle at E .

To Find : Length of AB

OP ⏊ PT [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

By phythagoras theorem in △OPT right angled at P

(OT)² = (OP)²+ (PT)²

(13)² = (5)² + (PT)²

(PT)²= 169 - 25 = 144

PT = 12 cm

PT = TQ = 12 cm [Tangents drawn from an external point to a circle are equal]

Now,

OT = OE + ET

ET = OT - OE = 13 - 5 = 8 cm

Now, as Tangents drawn from an external point to a circle are equal .

AE = PA [1]

EB = BQ [2]

Also OE ⏊ AB [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠AEO = 90°

∠AEO + ∠AET = 180° [By linear Pair]

∠AET = 90°

In △AET By Pythagoras Theorem

(AT)² = (AE)² + (ET)²

[Here AE = PA as tangents drawn from an external point to a circle are equal]

(PT - PA)² = (PA)² + (ET)²

(12 - PA)² = (PA)² + (8)² [from 1]

144 + (PA)² - 24PA = (PA)² + 64

24PA = 80

PA = 80/24 = 10/3 [3]

∠AET + ∠BET = 180 [Linear Pair]

90 + ∠BET = 180

∠BET = 90

In △BET, By Pythagoras Theorem

(BT)² = (BE)²+ (ET)²

(TQ - BQ)² = (BQ)² + (ET)² [from 2]

(12 - BQ)² = (BQ)² + (8)²

144 + (BQ)² - 24BQ = (BQ)² + 64

24BQ = 80

BQ = 80/24 = 10/3 [4]

So,

AB = AE + BE

AB = PA + BQ [From 1 and 2]

AB = 10/3 + 10/3 [From 3 and 4]

AB = 20/3 cm

Given,

∠TRQ = 30°.

At point R, OR ⊥ RQ.

So, ∠ORQ = 90°

⟹ ∠TRQ + ∠ORT = 90°

⟹ ∠ORT = 90°- 30° = 60°

It’s seen that, ST is diameter,

So, ∠SRT = 90° [ ∵ Angle in semicircle = 90°]

Then,

∠ORT + ∠SRO = 90°

∠SRO + ∠PRS = 90°

∴ ∠PRS = 90°- 30° = 60°

Correct answer is (a) 8 cm

In right triangle PTO 

By using Pythagoras theorem, we have

PO² = OT² + TP²

\(\Rightarrow\) 10² = 6² + TP²

\(\Rightarrow\) 100 = 36 + TP²

\(\Rightarrow\) TP² = 64

\(\Rightarrow\) TP = 8cm

Correct answer is (c) 25 cm

The tangent at any point of a circle is perpendicular to the radius at the point of contact

∴ OT ⊥ PT

From right – angled triangle PTO,

∴ OP² = OT² + PT²     [Using Pythagoras' theorem]

\(\Rightarrow\) OP² = (7)² + (24)²

\(\Rightarrow\) OP² = 49 + 576

\(\Rightarrow\) OP² = 625

\(\Rightarrow\) OP = \(\sqrt{625}\)

\(\Rightarrow\) OP = 25 cm