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Answer is (C) 110°

∵ ∠CBA + 70° = 180° (linear pair)

∠CBA = 180° – 70° = 110°

∵ ABCD is cyclic quadrilateral

We know that sum of opposite angles of cyclic quadrilateral is 180°

Thus, ∠CBA + ∠CDA = 180°

⇒ 110° + ∠CDA = 180°

⇒ ∠CDA = 180° – 110°

= 70°

But ∠x + ∠CDA = 180° (linear pair)

∠x + 70° = 180°

∠x = 180° – 70°

= 110°

From P we have tangents PA and PB

Hence PA = PB …tangents from same point are equal …(a)

Point Q is on PA

From Q we have tangents QA and QC

⇒ QA = QC …tangents from same point are equal …(i)

Point R is on PB

From R we have two tangents RC and RB

⇒ RC = RB … tangents from same point are equal …(ii)

Consider ΔPQR

⇒ perimeter of ΔPQR = PQ + QR + PR

From figure QR = QC + CR

⇒ perimeter of ΔPQR = PQ + QC + CR + PR

Using (i) and (ii)

⇒ perimeter of ΔPQR = PQ + QA + RB + PR

From figure we have

PQ + QA = PA and RB + PR = PB

⇒ perimeter of ΔPQR = PA + PB

Using (a)

⇒ perimeter of ΔPQR = PA + PA

⇒ perimeter of ΔPQR = 2(PA)

PA is 20 cm given

⇒ perimeter of ΔPQR = 2 × 20

⇒ perimeter of ΔPQR = 40 cm

Since, BO is the bisector of ∠ABC, then, 

∠ABO = ∠CBO …..(i) 

From figure: 

Radius of circle = OB = OA = OB = OC 

∠OAB = ∠OCB …..(ii) [opposite angles to equal sides] 

∠ABO = ∠DAB …..(iii) [opposite angles to equal sides] 

From equations (i), (ii) and (iii), we get 

∠OAB = ∠OCB …..(iv) 

In ΔOAB and ΔOCB: 

∠OAB = ∠OCB [From (iv)] 

OB = OB [Common] 

∠OBA = ∠OBC [Given] 

Then, By AAS condition : 

ΔOAB ≅ ΔOCB 

So, AB = BC [By CPCT]

Given that, 

BO is the bisector of ∠ABC 

To prove : 

AB = BC 

Proof : 

∠ABO = ∠CBO 

(BO bisector of ∠ABC) 

(i) OB = OA (Radii) 

Therefore, 

∠ABO = ∠DAB 

(Opposite angle to equal sides are equal) 

(ii) OB = OC (Radii) 

Therefore, 

∠CBO = ∠OCB 

(Opposite angles to equal sides are equal) 

(iii) Compare (i), (ii) and (iii) 

∠OAB = ∠OCB 

(iv) In triangle OAB and OCB, we have 

∠OAB = ∠OCB [From (iv)] 

∠OBA = ∠OBC (Given) 

OB = OB (Common) 

By AAS congruence rule,

Δ OAB ≅ Δ OCB 

AB = BC (By c.p.c.t)

Hence, proved

(1) ↔ (c), 

(2) ↔ (a), 

(3) ↔ (d), 

(4) ↔ (b)

Correct option  (a)  2 or –3/2

Explanation:

Condition for two circles to intersect at right angles is 2g₁ g₂ + 2f₁ f₂ = c₁ + c₂

Here two circles are x² + y² + 2x + 2ky + 6 = 0

and  x² + y² + 2ky + k = 0

g₁ = 1, f₁ = k c₁ = 6

g₂ = 0 f₂ = k c₂ = k

0 + 2k² = 6 + k

2k² – k – 6 = 0

2k² – 4k + 3k – 6 = 0

(2k + 3) (k – 2) = 0

k = –3/2 or k = 2

Given x² + y² + 6x – 2y – 6 = 0 & P (1, -2) centre = (-3, 1) 

Slope of the diameter = m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{1 - (-2)}{-3 - 1}\)= \(\frac{3}{-4}\)

∴ Equation of the diameter with (1, -2) & m = \(\frac{-3}{4}\)

y – y₁ = m(x – x₁) 

y + 2 = \(\frac{-3}{4}\)(x - 1) 

4y + 8 = -3x + 3 

⇒ 3x + 4y + 5 = 0.

Given x² + y² – 6x + 15y – 6 = 0 

C = \((3, \frac{-15}{2})\) & c = -16 

Length intercepted by x – axis = 2\(\sqrt{q^2 - c}\) 

\(= 2\sqrt{(3)^2 - (-16)}\) = 2\(\sqrt{9 + 16}\) = 2√25 = 10

∴ Length = 10 units

Given x² + y² – 6x – 4y – 12 = 0 

C = (3, 2) C = -12 

Length intercepted by x-axis \(2\sqrt{q^2 - c} \) = 2\(\sqrt{9 + 12}\)= 2√21 units 

Length intercepted by y-axis = \(2\sqrt{f^2 - c}\) = \(2\sqrt{4 + 12}\) = 2√16 = 8 units.

Given x² + y² + ax + by – 3 = 0 & center = (1, -3), a = ?, b = ? 

Here \(\frac{a}{2}\) = 1 & \(\frac{b}{2}\) = -3 

⇒ a = 2 & b = – 6.

PT is tangent to circle and OP is radius

⇒ ∠OPT = 90° …radius is perpendicular to tangent

From figure

⇒ ∠OPT = ∠OPB + ∠BPT

⇒ 90° = ∠OPB + 50° …∠BPT is 50° given

⇒ ∠OPB = 40°

Hence ∠OPB is 40°

=> Secant to a circle is a line which intersects the circle in two distinct points.

=> A tangent to a circle is a line that intersects the circle in exactly one point.

A circle is the locus of a points which moves in a plane in such a way that its distance from a fixed point remains constant.

‘The distance travelled when the wheel is rotated once =2 π r = 2 × π × 30

= 188.4 cm = 1.884 m.

The time required for the wheel to travel a distance of 200m = 200/1.884 = 106.16 = 106

Area of the rectangle = 24 × 14 = 336 cm²

Area of two semicircle = Area of a complete circle

Diameter of the circle = 14 cm

Radius = 7 cm

Area of the circle = π r²

= π × 72 = 49

π = 153.86 cm² Area of the remaining portion

= 336 – 153.86= 182.14 cm²

Area, of the rectangle = 50 × 20 = 1000 cm²

If the two semicircles cut out are joined it becomes a circle

Its radius = 20/2 = 10 cm

Area of the portion cut out

= πr² = π × 10 × 10 = 3.14 × 10 × 10 = 314 cm²

Area of the remaining portion

= 1000 – 314 = 686 cm²

PQ is tangent to circle and OP is radius

⇒ ∠OPQ= 90° …radius is perpendicular to tangent

⇒ ∠PQO = 30° …given

Consider ΔOPQ

⇒ ∠OPQ + ∠PQO + ∠POQ = 180° …sum of angles of triangle

⇒ 90° + 30° + ∠POQ = 180°

⇒ 120° + ∠POQ = 180°

⇒ ∠POQ = 60°

Hence ∠POQ is 60°

Consider ABCD as a parallelogram touching the circle at points P, Q, R and S as shown

As ABCD is a parallelogram opposites sides are equal

⇒ AB = CD …(a)

⇒ AD = BC …(b)

AP and AS are tangents from point A
⇒ AP = AS …tangents from point to a circle are equal…(i)

BP and BQ are tangents from point B
⇒ BP = BQ …tangents from point to a circle are equal…(ii)

CQ and CR are tangents from point C
⇒ CR = CQ …tangents from point to a circle are equal…(iii)

DR and DS are tangents from point D
⇒ DR = DS …tangents from point to a circle are equal…(iv)

Add equation (i) + (ii) + (iii) + (iv)

⇒ AP + BP + CR + DR = AS + DS + BQ + CQ

From figure AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC

⇒ AB + CD = AD + BC

Using (a) and (b)

⇒ AB + AB = AD + AD

⇒ 2AB = 2AD

⇒ AB = AD

AB and AD are adjacent sides of parallelogram which are equal hence parallelogram ABCD is a rhombus

Hence if all sides of a parallelogram touch a circle then that parallelogram is a rhombus.

i. Here, circle with centres A and B touch each other externally at point C.

∴ d(A, B) = d(A, C) + d(B ,C) 

= 3 + 4 

∴ d(A, B) = 7 cm

[The distance between the centres of circles touching externally is equal to the sum of their radii] 

ii. Here, circle with centres A and 13 touch each other internally at point C.

[The distance between the centres of circles touching internally is equal to the difference in their radii]

From P we have two tangents PA and PB

The tangents from an external point to a circle are equal

Hence PA = PB

PA = 6 cm given

Hence PB is also 6 cm

Circles with centres N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K – N – M]. 

Hence, the given circles are internally touching circles.

By using angle sum property in triangle ABC, 

∠B = 180° – (60° + 20°) = 100° 

In cyclic quadrilateral ABCD, we have 

∠B + ∠D = 180° 

∠D = 180° – 100° 

= 100°

(C) The angles A and C is 125⁰