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:(a)x+y=2(b)x-y=2(c)2x-y=1(d)none of theseI WILL B VERY THANKFUL TO U IF U GIVE AN ELABORATED SOLUTION...THANKU

let one of the line is y=mx                   (line is passing through origin)

this line makes an angle @ with x axis ...another line is also inclined to x axis with same angle but its slope will be -m..

now other will be y=-mx

let point of contact of line y=mx & y=-mx with circle be P & q respectively then

on solving y=mx and circle we get

P=[  2(1+m)/1+m²   ,   2m(1+m)/1+m²  ]

and on solving y=-mx and circle we get

q=[  2(1-m)/1+m²   ,   2m(m-1)/1+m²   ]

we have to find the locus of mid point of PQ because pq is chord of circle,

let X,Y is the mid point then

            X=2/1+m²  &  Y=2m² /1+m²

on solving these we get

                X+Y=2

THIS IS THE REQUIRED EQUATION