A circle S ≡ 0 passes through the common points of family of cir

1.	A circle S ≡ 0 passes through the common points of family of circles x2 + y2 + λx – 4y + 3 = 0(λ ∈ R) and have minimum area then(a) area of S ≡ 0 is π sq. u(b) radius of director circle of S ≡ 0 is √2(c) Radius of director circle of S ≡ 0 for x–axis is 1 unit(d) S ≡ 0 never cuts \|2x\| = 1
Answer» 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

A circle S ≡ 0 passes through the common points of family of circles x2 + y2 + λx – 4y + 3 = 0(λ ∈ R) and have minimum area then(a) area of S ≡ 0 is π sq. u(b) radius of director circle of S ≡ 0 is √2(c) Radius of director circle of S ≡ 0 for x–axis is 1 unit(d) S ≡ 0 never cuts |2x| = 1

Answer»

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