The pair of straight lines joining the origin to the points of intersection of the line `y=2sqrt(2x)+c` and the circle `x^2+y^2=2`1 are at right angles , if

Options

A. `c^2-4=0`
B. `c^2-8=0`
C. `c^2-9=0`
D. `c^2-10=0`

Answer

C

Explanation

Given , equation of circle3 is `x^2+y^2=2`.
and equation of line is `((y-2sqrt(2x))/(c))=1`
Making Eq. (i) homogeneous with the help of Eq.(ii) , we get `x^2+y^2-2((y-2sqrt(2)x)/(c))^2=0`
`rArr x^2+y^2-(2)/(c^2)(y^2+8x^2-4sqrt(2)xy)=0`
`rArr (1-(16)/(c^2))x^2+(1-(2)/(c^2))y^2+(8sqrt(2))/(c^2)xy=0` ....(iii)
The lines represented by Eq. (iii) will be perpendicular , if `1-(16)/(c^2)+1-(2)/(c^2)=0`
`rArr 2-(18)/(c^2)=0" " [because "Condition for perpendicular is "a+b=0]`
`rArr 2c^2=18rArr c^2=9`
`therefore c^2-9=0`.