If the lines joining the origin to the points of intersection of the line `y=mx+2` and the curve `x^(2)+y^(2)=1` are at right-angles, then

Options

A. `m=sqrt(3)`
B. `m=+-sqrt(7)`
C. `m=1`
D. `m=sqrt(5)`

Answer

B

Explanation

We have, `x^(2)+y^(2)=1`
`implies`Centre=(0,0) and radius=1

Since, lines are intersecting at right angles.
i.e., AB is the diameter of another circle which passes through (0,0)
and have radius `=(sqrt(2))/(2)=(1)/(sqrt(2))`
`therefore` Distance of line `mx-y+2=0`
from (0,0),
`(1)/(sqrt(2))=(|m*0-0+2|)/(sqrt(m^(2)+1))`
`implies(m^(2)+1)=2*2*2`
`impliesm^(2)=7`
`thereforem=+-sqrt(7)`