Find the equations of straight lines which pass through the intersection of the lines `x -2y-5=0, 7x +y =50` & divide the circumference of the circle `x^2 + y^2 =100` into two arcs whose lengths are in the ratio 2:1.

Options

A. `3x+4y-25=0`
B. `4x-3y-25=0`
C. `3x+2y-23=0`
D. `2x-3y-11=0`

Answer

1,2

Explanation

Point of intersection of given lines is `(7,1)`, which lies inside the circle `x^(2)+y^(2)=100`
If line chord AB thought this point divides circumeference of the circle into two arcs whose lengths are in the ratio `2:1`, then AB subtends an angle of `120^(@)` at the centre
So, from the figure, `OM = 10cos60^(@) = 5`
Let the equation of line through `(7,1)` be
`y-1 = m (x-7)`
or `mx-y-7m+1=0` (1)
`OM =5`
`implies |(1-7m)/(sqrt(1+m^(2)))|=5`
`implies (1-7m)^(2)= 25 ( 1+m^(2))`
`implies 24m^(2)-14m-24=0`
`implies (4m+3)(6m-8)=0`
`implies m= -(3)/(4), m= (4)/(3)`
Putting the value of m in (i) , we get
`y-1= - (3)/(4)(x-7) ` and `y-1= (4)/(3)(x-7)`
or `3x +4y - 25 =0` and `4x-3y-25=0`