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A circle S= 0 passes through the common ...

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

A. area of `S = 0` is `pi` sq. units

B

B. radius of director circle of `S = 0` is `sqrt(2)`

C

C. radius of director circle of `S = 0` is 1 unit

D

D. `S = 0` never cuts `|2x| =1`

Text Solution

Verified by Experts

The correct Answer is:
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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