A circle `S= 0` passes through the common points of family of circles `x^2 +y^2 +lambdax-4y +3=0` and `(lambda epsilon R)` has minimum area then (A) area of `S = 0` is `pi` sq. units (C) radius of director circle of `S = 0` is `1` unit (D) `S = 0` never cuts `|2x|=1` (B) radius of director circle of `S = 0` is `sqrt2`A. area of `S = 0` is `pi` sq. unitsB. radius of director circle of `S = 0` is `sqrt(2)`C. radius of director circle of `S = 0` is 1 unitD. `S = 0` never cuts `|2x| =1`

A circle `S= 0` passes through the common points of family of circles

1.	A circle `S= 0` passes through the common points of family of circles `x^2 +y^2 +lambdax-4y +3=0` and `(lambda epsilon R)` has minimum area then (A) area of `S = 0` is `pi` sq. units (C) radius of director circle of `S = 0` is `1` unit (D) `S = 0` never cuts `\|2x\|=1` (B) radius of director circle of `S = 0` is `sqrt2`A. area of `S = 0` is `pi` sq. unitsB. radius of director circle of `S = 0` is `sqrt(2)`C. radius of director circle of `S = 0` is 1 unitD. `S = 0` never cuts `\|2x\| =1`
Answer» Correct Answer - A::B::D `x^(2) +y^(2) + lambda x - 4y +3 = 0` `:. x^(2) + y^(2) - 4y +3 + lambda x = 0` Common points on circle are point of intersection of `x = 0` and `y^(2) - 4y +3 = 0` So common points are `A(0,1)` and `B(0,3)`. `:.` Circle has minimum area if AB is diameter `:.` Equation of circle is `x^(2) +y^(2) -4y +3 = 0` So radius of director circle is `sqrt(2)`. Area of circle `S = 0` is `pi` sq. units. Also `\|2x\| =1` never cuts circle.

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