semicircle

[bug]

manish, gh - would you be able to take a look at this q please.


thx

[catchkedar]

Point(-sqrt3 ,1 ) & (s,t) are on circle.
Slope of PO = -1/sqrt3
So Slope of OQ = sqrt3 ....as both segments are perpendicular to each other product of slope is -1
Equation of line OQ is t=(sqrt3)*s ......equn of line is y=mx+c here y is t,x is s & m which is slope is sqrt3 ------Equn 1

Also as (s,t) point is on circle it follows circle equation x^2 + y^2 =a^2 where a is radius of circle

Find radius from point given on circle which is P....so radius of circle is
(sqrt3)^2 + 1^2 = radius^2
=3+1
=4
So radius =2
Now use this with new point on circle Q
So s^2 + t^2 =2^2=4
Substitute value of t from Equn 1 above
s^2 + 3s^2 =4
4s^2=4
So s=1
Hence Answer is B........

[bug]

thnx a lot catchkedar. neat explanation.
btw, the answer is B

[GMAT-HELP]

http://www.urch.com/forums/gmat-math...rep-trick.html (2 GMAT PREP, Trick)