If the circumference of the circle `x^2+y^2+8x+8y-b=0` is bisected by the circle `x^2+y^2=4` and the line `2x+y=1` and having minimum possible radius is

Options

A. a)`5x^(2)+5y^(2)+18x+6y-5=0`
B. b)`5x^(2)+5y^(2)+9x+8y-15=0`
C. c)`5x^(2)+5y^(2)+4x+9y-5=0`
D. d)`5x^(2)+5y^(2)-4x-2y-18=0`

Answer

4

Explanation

Let the equation of the circle be
`x^(2)+y^(2)-4+k(2x+y-1)=0`
where k is a real number
Radius `=sqrt((5k^(2))/(4)+4+k)`
Radius is minimum when `k= - 2//5`. ltbr. Therefore, the required equation will be
`5x^(2)+5y^(2)-4x-2y-18=0`