Slope of line PO = -1/sqrt(3)
because two points are (0,0) and (-sqrt(3), 1)
Also radius of circle from two points (0,0) and (-sqrt(3), 1) is 2.
equn of circle x^2+y^2 = 4
Line OQ is perpendicular to Op
so slope = sqrt(3)
and equn => y = sqrt(3) * x
Now we have two equations at point Q
s^2 + t^2 = 4
and
t = sqrt(s)
Solve you will get your answer.
I am sure there must be some shortcut also, but as of now I can not think of.
A request: If possible please type the question. Lot of us will be able to save some time. Thanks a lot.
