Let `L_1` be a straight line passing through the origin and ` L_2` be the straight line `x + y = 1` if the intercepts made by the circle `x^2 + y^2-x+ 3y = 0` on `L_1` and `L_2` are equal, then which of the following equations can represent `L_1`?

Options

A. `x+y=0, x-7y=0`
B. `x-y=0, x+7y=0`
C. `7x+y=0`
D. `x-7y=0`

Answer

B

Explanation

The centre and radius of the given circle are `((1)/(2), -(3)/(2))` and `sqrt((5)/(2))` respectively.
Let `y=mx` be the equation of `L_(1)`. Then,
`p_(1)`= Length of the intercept on `L_(1)`
`rArr p_(1)=2sqrt((sqrt((5)/(2)))^(2)-((m+3)/(2sqrt(m^(2)+1)))^(2))=2sqrt((5)/(2)-((m+3)^(2))/(4(m^(2)+1)))`
and,
`rArr p_(2)=` Length of the intercept on `L_(2)`
`rArr p_(2)=2sqrt((sqrt((5)/(2)))^(2)-(sqrt(2))^(2))=2sqrt((5)/(2)-2)=sqrt(2)`
Now,
`p_(1)=p_(2)`
`rArr 2sqrt((5)/(2)-((m+3)^(2))/(4(m^(2)+1)))=sqrt(2)`
`rArr 5-((m+3)^(2))/(4(m^(2)+1))=1`
`rArr 7m^(2)-6m-1=0`
`rArr (m-1)(7m+1)=0rArr m=1, -(1)/(7)`
So, the equations of `L_(1)` are `y=x` and `7y=-x`.