The lines joining the point of intersection of the line x+y=1 and the curve `x^2+y^2-2y+lambda=0` to the origin are perpendicular, then value of `lambda` wil be:

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. `(1)/(sqrt(2))`
D. 0

Answer

D

Explanation

Given , equation of curve is `x^2+y^2-2y+lambda=0` ....(i)
and equation of line is `x+y=1` ......(ii)
Making Eq. (i) homogeneous with the help of Eq. (ii) , we get `x^2+y^2-2y(x+y)+lambda(x+y)^2=0`
`rArr x^2+y^2-2xy-2y^2+lambda(x^2+y^2+2xy)=0`
`rArr (1+lambda)x^2+(lambda-1)y^2+(lambda-1)2xy=0` ....(iii)
The lines represented by Eq. (iii) are perpendicular. `therefore (1+lambda)+lambda -1=0`
`rArr lambda=0 [because "condition for perpendicular "a+b=0]`.