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Given: ∠BOC = 100° 

By degree measure theorem: ∠AOC = 2 ∠APC 

100° = 2 ∠APC 

or ∠APC = 50° 

Again, ∠APC + ∠ABC = 180° (Opposite angles of a cyclic quadrilateral) 

50° + ∠ABC = 180° 

or ∠ABC = 130° 

Now, ∠ABC + ∠CBD = 180° (Linear pair) 

130° + ∠CBD = 180° 

or ∠CBD = 50°

Correct option is (C) Rectangle

A cyclic parallelogram is a rectangle.

C) Rectangle

Correct option is (B) Right angle

Angle in a semi-circle is a right angle.

Correct option is  B) Right angle

Given: ∠OBD = 50° 

Here, AB and CD are the diameters of the circles with centre O. 

∠DBC = 90° ….(i) [Angle in the semi-circle] 

Also, ∠DBC = 50° + ∠OBC 

90° = 50° + ∠OBC 

or ∠OBC = 40° 

Again, By degree measure theorem: ∠AOC = 2 ∠ABC 

∠AOC = 2∠OBC = 2 x 40° = 80°

Given: m(∠CAB)= 30° 

To Find: m(∠ACB) and m(∠ABC). 

Now, ∠ACB = 90° (Angle in semi-circle) 

Now, 

In △ABC, by angle sum property: 

∠CAB + ∠ACB + ∠ABC = 180° 

30° + 90° + ∠ABC = 180° 

∠ABC = 60° 

Answer: ∠ACB = 90° and ∠ABC = 60°

Data: In ∆ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect each other. O is the centre of the circle. 

To Prove: Angle bisector of ∠A and perpendicular bisector of BC intersect at D. 

Construction: Join OB, OC. 

Proof: Angle subtenced at the Centre 

= 2 × angle subtended in the circumference. 

∠BOC = 2 × ∠BAC 

In ∆BOE and ∆COE, 

∠OEB = ∠OEC = 90° (∵ OE⊥BC) 

∴ BO = OC (radii) OE is common. 

∴ ∆BOE ≅ ∆COE (RHS postulate) 

But, ∠BOE + ∠COE = ∠BOC 

∠BOE + ∠BOE = ∠BOC 

2∠BOE = ∠BOC 

2∠BOE = 2∠BAC 

∴ ∠BOE = ∠BAC 

But, ∠BOE = ∠COE = ∠BAC 

∠BAD = ∠BAC 

∠BAD = ∠BOE 

∠BAD = ∠BOD 

∴ ∠BOD = 2∠BAD 

∴ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference. 

∴ Angle bisector of ∠A and perpendicular bisector of BC intersect at D.

By degree measure theorem: ∠AOB = 2 ∠APB 

so, ∠AOB = 2 × 70° = 140° 

Since AOBC is a cyclic quadrilateral, we have 

∠ACB + ∠AOB = 180°

∠ACB + 140° = 180° 

∠ACB = 40°

(A) The angles CBD is 25⁰.

The correct answer is (C) 80⁰  

Suppose two circles intersect in three points A, B and C. 

Then,

 A, B, C are non-collinear. 

So, 

A unique circle passes through these three points. 

This is contradiction to the face that two given circles are passing through A, B, C. 

Hence, 

Two circles cannot intersect each other at more than two points.

Radius of the circle is 3 cm and PQ = 4.8 cm.

PQ is a chord of the circle with centre O.

The tangents at P and Q intersect at point T

So, TP and TQ are tangents

OP and OT are joined.

Join OQ

We have two right triangles: △OPT and △OQT

Here,

OT = OT (Common)

PT = QT (tangents of the circle)

OP = OQ (radius of the same circle)

By Side – Side – Side Criterion

△OPT ≅△OQT

Therefore, ∠POT = ∠OQT

Again, from triangles △OPR and △OQR

OR = OR (Common)

OP = OQ (radius of the same circle)

∠POR = ∠OQR (from above result)

By Side – Angle – Side Criterion

△OPR ≅△OQT

Therefore, ∠ORP = ∠ORQ

Now,

∠ORP + ∠ORQ = 180°

(Sum of linear angles = 180 degrees)

∠ORP + ∠ORP = 180°

∠ORP = 90°

This implies, OR ⏊ PQ and RT ⏊ PQ

Also OR perpendicular from center to a chord bisects the chord,

PR = QR = PQ/2 = 4.8/2 = 2.4 cm

Applying Pythagoras Theorem on right triangle △OPR,

(OP)² = (OR)² + (PR)²

(3)² = (OR)² + (2.4)²

OR = 1.8 cm

Applying Pythagoras Theorem on right angled △TPR,

(PT)² = (PR)² + (TR)² …(1)

Also, OP ⏊ OT

Applying Pythagoras Theorem on right △OPT,

(PT)² + (OP)² = (OT)²

[(PR)² + (TR)² ] + (OP)² = (TR + OR)²

(Using equation (1) and from figure)

(2.4)² + (TR)² + (3)² = (TR + 1.8)²

4.76 + (TR)² + 9 = (TR)² + 2(1.8)TR + (1.8)²

13.76 = 3.6 TR + 3.24

TR = 2.9 cm [approx.]

From (1) => PT² = (2.4)² + (2.9)²

PT² = 4.76 + 8.41

or PT = 3.63 cm [approx.]

Answer: Length of PT is 3.63 cm.

Equation of tangent of x² + y² – 2ax = 0 at (a(1 + cosα),asinα) is 

ax(1 + cosα) + aysin α – a(x + a(1 + cosα)) = 0

ax + axcosα + aysinα – ax – a² (1+cosα) = 0

axcosα + aysinα = a² (1 + cosα)

xcosα + ysinα = a(1 + cosα) 

Since tangents make an angle of 60° with the x–axis so slope of tangent

m = tan 60° = m = √3

radius of circle x² + y² = 9 is 3

we know equation of tangent to a circle x² + y² = a² is

y = mx ± a√1 + m²

∴ y = √3x ± 3 √1 + 3

= √3x ± 6

or √3x – y ± 6 = 0 ie., √3x - y - 6 = 0  are equations of tangents.

Since the line (x – 2) cosθ + (y – 2) sinθ = 1 .......(1)

touches a circle so it is a tangent equation to a circle.

Equation of tangent to a circle at (x₁ ,y₁ ) is (x – h)x₁ +(y – k)y₁

= a² to a circle (x – y)² + (y – k)² =  a² comparing (1) and (2) we get

x – h = x – 2 y – k = y – 2 and a² = 1

x₁ = 1cosθ  y₁ = 1sinθ

∴ Required equation of circle is

(x – 2)² + (y – 2)² = 1

x² + y² – 4x – 4y + 7 = 0

Equation of normal at (x₁ ,y₁ ) of x² + y² – 2x = 0 is

x - x₁/x₁ - 1 = y - y₁/y₁ - 0 (∴ x - x₁/ax₁ + g = y - y₁/by₁ + f)

Slope of this equation is y₁/x₁ - 1

Slope of x + 2y = 3 is -1/2

Since given that normal is parallel to x + 2y = 3

∴  y₁/x₁ - 1 ₌ -1/2

y₁ = – x₁ + 1 therefore locus of (x₁ , y₁ ) is x₁ + 2y₁ = 1

It is the equation of normal

Given circles are

S₁ = x² + y² + 4x + 2y + 1 = 0

S₂ = x² + y² – 2x + 6y – 6  = 0

S₁ –S₂ = 0

6x – 4y + 7 = 0

System of co-axial circle is S₁ + λ(S₁ –S₂) = 0

x² + y² + 4x + 2y + 1 + λ(6x – 4y + 7) = 0

x² + y² + 2x(2 + 3λ) + 2y(1 – 2λ) + 1 + 7λ = 0

Centre of this circle is (–(2 + 3λ), – (1–2λ)

lies on radical axis

6( – 2 – 3λ) + 4(1 – 2λ) + 7 = 0 

– 12 – 18λ + 4 – 8λ + 7

=  –1 –26λ = 0

λ = -1/26

∴ Required particular member of co-axial circle is 26(x² + y²) + 98x + 56y + 19 = 0

(c) 125°

∠AOB + ∠APB=180° (The angle between two tangents drawn from an external pt. to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre) 

⇒ ∠AOB= 180° – 70°= 110° 

Reflex ∠AOB= 360° – 110° = 250° 

∴ ∠BPC = \(\frac{1}{2}\)× Reflex ∠AOB = \(\frac{1}{2}\) × 250° = 125°

False

Justification:

A circle through two points cannot pass through a point which is collinear to these two points.

Hence, the given statement is false.

True

Justification:

According to the question,

Radius of circle = 3 cm

Diameter of circle = 2 × r

= 2 × 3 cm

= 6 cm

Now,

From the question we have,

AB = 6 cm

So, the given statement is true because AB will be the diameter

(c) 68°

∠BOC = 2 × ∠BAC = 2 × 22° = 44° 

(Angle at the centre = 2 × angle at any point on the remaining part of the circle)

∠AOC = 180° – ∠BOC = 180°– 44 = 136° (∵ AOB is a straight line)

\(\Rightarrow\) ∠ADC = \(\frac{1}{2}\) x ∠AOC = \(\frac{1}{2}\) x 136º = 68º 

(Angle at any pt. on the remaining part of the circle = \(\frac{1}{2}\) × angle at the centre)

(c) 40°

∠ADB =90° (Angle in a semi-circle) 

In ΔADB,∠DAB= 180°– (∠ADB+ ∠DBA) 

= 180° – (90° + 35°) = 180° – 125° = 55° (Angle sum prop. of a Δ)

In cyclic quad ABCD, 

∠A + ∠C = 180° ⇒∠C= 180° – 55° = 125° (Opp. ∠s of a cyclic quad. are supp.) 

∴ In ΔDCB,∠CBD = 180° – (15° + 125°) = 40°

(c) 108°

∠ACE = 180° – 54° = 126° (BCE is a straight line) 

⇒ ∠ADE = ∠ACE = 126° (Angles in the same segement are equal) 

∴ Reflex ∠AOE = 2 × ∠ADE = 2 × 126° = 252° (Angle at the centre = 2 × Angle at any other pt. on remaining part of the circle) 

∴ x = 360° – 252° = 108°.

(d) 21°

∠BAD = 90° (Angle in a semi-circle) 

∠CAD = \(\frac{1}{2}\)  x ∠COD =(Angle at the centre = 2× angle at any other point on the remaining part of the circle) 

∴ ∠BAC = ∠BAD – ∠CAD = 90°– 46° = 44° 

In ΔABE, ∠AEB = 180° – (65° + 44°) 

= 180° – 109° = 71° 

⇒ ∠CEO = ∠AEB = 71° (Vert. opp. ∠s) 

Also, ∠COE = 180° – 92° = 88° (BOD is a st. line)

∴ In ΔCEO, y = ∠ECO = 180° – (88° + 71°) 

= 180° – 159° = 21°.

(b) 35°, 62°

∠DAC = ∠DBC = 45° (Angles in same segment) 

∠DAB = 80° (Ext. ∠ of a cyclic quad = int. opp. ∠) 

∴ x = ∠DAB – ∠DAC= 80° – 45° = 35° 

Now,∠ADB = \(\frac{1}{2}\)×∠AOB = \(\frac{1}{2}\)× 76°= 38°  

∠ACB = ∠ADB = 38° (Angles in the same segment)

∴ y = 180° – (38° + 80°) =180° – 118° = 62° ( ∵DLF is a st. line)

As arc AXB = 1/2 arc BYC, 

∠AOB = 1/2 ∠BOC 

Also ∠AOB + ∠BOC = 180º 

Therefore, 1/2 ∠BOC + ∠BOC = 180º

or ∠BOC = 2/3 x 180º = 120º

a1=2a+b
a2 =4a+b
a3=6a+b
Common difference = d=( 4a + b ) - ( 2a + b ) = 2a

Therefore, sum of first k terms =k/2[(2a+(k-1)d]= k/2[(2(2a+b)+(k-1)2a]= k/2 x 2
(2a+b+k-a)=k(a+b+ak)

Solution: a = 6 , d = 6 and n = 40

S₄₀ = 40/2 (12 + 6 x 39) =20 x 336 = 6720

(d) A circle can have more than two parallel tangents. parallel to a given line.

A circle can have more than two parallel tangents. parallel to a given line. This statement is false because there can only be two parallel tangents to the given line in a circle.

(d) A tangent to the circle can be drawn form a point inside the circle.

A tangent to the circle can be drawn from a point Inside the circle. This statement is false because tangents are the lines drawn from an external point to the circle that touch the circle at one point.

(d) A straight line can meet a circle at one point only

A straight be can meet a circle at one point only 

This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.

(d) A straight line can meet a circle at one point only.

A straight be can meet a circle at one point only This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.

‘a’ is minor segment and ‘b’ is major segment.