The straight line `x+y-1=0` meets the circle . `x^2 +y^2-6x-8y=0` at A and B. Then, the equation of : the circle of which AB is a diameter, is

Options

A. `x^(2)+y^(2)-2y-6=0`
B. `x^(2)+y^(2)-2y-6=0`
C. `2(x^(2)+y^(2))+2y-6=0`
D. `3(x^(2)+y^(2))+2y-6=0`

Answer

B

Explanation

Required equation of cricle is `x^(2)+y^(2)-6x-8y+lamda(x+y-1)=0` or x^(2)+y^(2)-(6-lamda)x-(8lamda)y-lamda=0` whose centre is `(3-lamda/2+4-lamda/2-1=0` and which lies on the line x+y-1=0.
rArr 3-lamda/2+4-lamda/2-1=0`
`rArr lamda=6` hence, required equation is `x^(2) +y^(2)-6x-8y+6x+6y-6=0`
`rArrx^(2)+y^(2)-2y-6=0`