Circle Question

[antyagi]

I could not find the solution of the question attached.
This is from GMATPrep

[sudhir3127]

OP = OQ ( assume O has coodinates(0,0) )

Distance betwwen two points

(1-0)^2 + (-sqrt3-0)^2 = (s-0)^2 + (t-0)^2
1+3 = s^2+t^2

s^2+t^2 = 4 ---------(1)


Since the lines are perpendicular to each other, product of slopes = -1

(Y2-Y1)/(X2-X1) * (Y2-Y1)/(X2-X1)= -1

[(1-0)/(-sqrt3-0)]*[(t-0)/(s-0)] = -1
t/s = sqrt3
t = s*sqrt3
From (1), s^2 + 3s^2 = 4
s^2 = 1

AS S is in 1st quadrant s=1

hope this helps.

[antyagi]

Thanks a lot.

[rosh26]

Is there another way to get this solution besides the distance formula? Can we use properties od triangles?

[rosh26]

Is there another way to get this solution besides the distance formula? Can we use properties od triangles?

[Ian Stewart]

If you understand what it means for slopes to be perpendicular, the question can be solved almost immediately. I explained how here:

www.beatthegmat.com/gmat-prep-geometry-t13546.html

Stuart also posted a very clear explanation of this approach in an earlier thread- follow the link from the thread above.

Or, you can notice that you get 30-60-90 triangles if you connect the point P to the x-axis and work from there; that approach is explained in the above thread as well.