Circle and Conic Section

6 )

One of the diameters of the circle circumscribing the rectangle ABCD is x – 4y + 7 = 0. If A and B are points (– 3, 4) and (5, 4) respectively, then the area of the rectangle is ……..

64 sq. units

8 sq. units

32 sq. units

16 sq. units

View Solution

Answer :

Answer: (a) 32 sq. units

Diameter is x – 4y + 7 = 0

point A(– 3, 4) & B(5, 4) does not lie on the diameter.

Let E be centre of circle.

∵ ABCD is a rectangle

|EA| = |EB| and E(a, b)

∴ (a + 3)² + (b – 4)² = (a – 5)² + (b – 4)²

∴ 6a + 9 = – 10a + 25

∴ a = 1

and E lie on line x – 4y + 7 = 0

⇒ a – 4b = 7 ⇒ b = 2

∴ E(a, b) = (1, 2)

Now |AB| = √[(5 + 3)² + (4 – 4)²] = 8

|BD| = 2|EB|

|BD| = 2√[(5 – 1)² + (4 – 2)²] = 2√(20) = 4√5

From ΔABD,

AD² = BD² – AB² = (4√5)² – (8)²

= 80 – 64

= 16

∴ |AD| = 4

∴ Area of rectangle ABCD = |AB| |AD|

= (8)(4)

= 32 sq. units.

7 )

The line x sin θ – y cos θ = k touches the circle x² + y² = k² then θ = ………

Î¸ âˆˆ [{(â€“ 1) / 2}, (1/2)]

Î¸ âˆˆ [0, 1]

Î¸ [â€“ 1, 1]

Î¸ is any angle

View Solution

Answer :

y = mx + c touches the circle if c² = a²(1 + m²)

given: y cos θ = x sin θ – k

∴ y = x tan θ – k sec θ

hence c² = k² sec² θ

and a²(1 + m²) = k²(1 + tan² θ) ---------- a² = k²

= k² sec² θ

∴ c² = a²(1 + m²)

⇒ θ is any angle as it is true for all values θ.

8 )

The radius of the circle passing through the points (5, 2), (5, – 2) and (1, 2) is

2âˆš2

5âˆš2

3âˆš2

2âˆš5

View Solution

Answer :

AB = √(4² + 0²) = 4

AC = √(4² + 4²) = 4√2

BC = √(0² + 4²) = 4

∴ AB² + BC² = AC²

i.e. 16 + 16 = 32

∴ It is right angled triangle.

∴ Diameter = ℓ(AC)

Diameter = 4√2

∴ Radius = 2√2.

9 )

Two tangents to the circle x² + y² = 4 at the points A and B meet at P(– 4, 0). The area of the quadrilateral PAOB, where O is the origin is

4âˆš3

6âˆš2

2âˆš3

View Solution

Answer :

circle: x² + y² = 4

i.e. radius = 2

sin θ = (2/4) = (1/2) ⇒ θ = 30°

∴ area of ΔPOA = (1/2) × 2 × 4 × cos θ

= 4 cos θ

= 4 cos 30

= 2√3

Area of quadrilateral

PAOB = 2 × (area of ΔPOA)

= 2 × 2√3

= 4√3.

10 )

If the tangents are drawn to the circle x² + y² = 12 at the point where it meets the circle x² + y² – 5x + 3y – 2 = 0, then the point of intersection of these tangent is

(6, â€“ 6)

[6, (18/5)]

[â€“ 6, (18/5)]

[6, (18/5)]

View Solution

Answer :

Let (h, k) be the point of intersection of tangents.

Then the chord of contact of tangent is the common chord of the circle

x² + y² = 12 and x² + y² – 5x + 3y – 2 = 0

i.e. 12 – 5x + 3y – 2 = 0 ⇒ 5x – 3y – 10 = 0 (1)

also equation of chord of contact is hx + ky – 12 = 0 (2)

Equation (1) & (2) represent the same line

∴ (h/5) = [k / (– 3)] = [(– 12) / 10]

⇒ h = – 6, k = (36 / 10) = (18 / 5)

∴ point of intersection of tangent (h, k) = [– 6, (18/5)].

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