The equation of the locus of the middle point of a chord of the circle `x^2+y^2=2(x+y)` such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is `x+y=2` (b) `x-y=2` `2x-y=1` (d) none of these

Options

A. x+y=2
B. `x-y=2`
C. `2x-y=1`
D. none of these

Answer

A

Explanation

Let (h, k) be the mid-point of a chord of the circle `x^(2)+y^(2)-2x-2y=0`. Then, the equation of the chord is
`hx+ky-(x+h) (y | k) = h^(2) |k^(2) 2h 2k`
`rArr x(h-1)+y(k-1)=h^(2)+k^(2)-h-k " " ...(i)`
The combined equation of the straight lines joining the origin to the points of intersection of the circle and the chord (i) is
`x^(2)+y^(2)-2(x+y) {(x(h-1)+y(k-1))/(h^(2)+k^(2)-h-k)}=0`
`rArr x^(2)(h^(2)+k^(2)-3h-k+2)+y^(2)(h^(2)+k^(2)-h-3k+2)-2xy(h+k-2)=0`
Lines represented by this equation are equally inclined to the x-axis.
`:.` Sum of their slopes = 0
`rArr (2(h+k-2))/(h^(2)+k^(2)-h-3k+2)=0 rArr h+k-2=0 rArr h + k =2`
Hence, the locus of (h, k) is x+y=2.