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

1

Explanation

Solving `y=mx` and `x^(2)+y^(2)-2x-2y=0`, we get
`x^(2)+m^(2)x^(2)-2x-2mx=0`
or `x=0,(2(1+m))/(1+m^(2))`

Similarly, solving `y= -mx` and the equation of the circle , we get
`x=0,(2(1-m))/(1+m^(2))`
`:gt A-=((2(1+m))/(1+m^(2)),(2m(1+m))/(1+m^(2)))`
and `B -=((2(1-m))/(1+m^(2)),(-2m(1-m))/(1+m^(2)))`
Let the middle point of AB be `(alpha,beta)`. Then
`alpha=(1)/(2){(2(1+m))/(1+m^(2)),(-2(1-m))/(1+m^(2))}`
and `beta=(1)/(2){(2m(1+m))/(1+m^(2)),(-2(1-m))/(1+m^(2))}`
`:. alpha=(2)/(1+m^(2)),beta=(2m^(2))/(1+m^(2))`
Eliminating m from these, we get `alpha+beta=2`.
Hence, the locus is `x+y=2`.