Correct Answer - 3
Let the equation of the chord OA of the circle
`x^(2)+y^(2)-2x+4y=0` (1)
by `y=mx` (2)
Solving (1) and (2) , we get
`x^(2)+m^(2)x^(2)-2x+4x=0`
or `(1+m^(2))x^(2)-(2-4m)x=0`
or `x=0` and `x=(2-4m)/(1+m^(2))`
Hence, the points of intersection are
`O(0,0)` and `A((2-4m)/(1+m^(2)),(m(2-4m))/(1+m^(2)))`
or `OA^(2)=((2-4m)/(1+m^(2)))(1+m^(2))=((2-4m)^(2))/(1+m^(2))`
Since OAB is an isosceles right-angles triagnle,
`OA^(2)=(1)/(2)AB^(2)`
where AB is a diameter of the given circle. Hence,
`OA^(2)=10`
or `((2-4m)^(2))/(1+m^(2))=10`
or `4-16m+16m^(2)=10+10m^(2)`
or `3m^(2)-8m-3=0`
i.e., `m=3` or `-(1)/(3)`
Hence, the required equations are `y=3x` or `x+3y=0`