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The circle S -= x(2) + y(2) + kx + (k + ...

The circle `S -= x_(2) + y_(2) + kx + (k + 1)y – (k + 1) = 0` always passes through two fixed point for every real k. If the minimum value of the radius of the circle S is `1/(sqrtP)` then the value of 'P' is :

A

4

B

6

C

8

D

16

Text Solution

Verified by Experts

The correct Answer is:
C
`x^(2)+y^(2)+y-1+k(x+y-1)=0`

`c(-k/2,-((k+1)/(2)))`
centre of required circle lie on AB
`implies -k/2-((k+1)/(2))-1=0`
`-2k -1 -2 = 0`
`k = -3/2`
`x^(2)+y^(2)+y-1-3/2(x+y-1)=0`
`implies 2x^(2)+2y^(2)+2y-2-3x-3y+3 =0`
`implies 2x^(2)+2y^(2)-3x-y+1=0`
`r = sqrt(9/16 + 1/16 -1/2) = sqrt(1/8)`
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