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reply2spg
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Prep Question...Need help 04 Jan 2010, 17:40
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xcusemeplz2009
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Prep Question...Need help 04 Jan 2010, 22:42
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OP=OQ=radius
OP=sqrt[(0-rut3)^2+(0-1)^2]=2-------------dis formula
radius=2---->OQ=2

OQ=sqrt[s^2+t^2]=4----->s^2+t^2=2....eqn 1

pythagoras theroem
OP^2+OQ^2=PQ^2

PQ^2= (s+RUT3)^2+(t-1)^2....DIS FORM
---->(s+RUT3)^2+(t-1)^2=4+4.........EQN 2
USING EQN 1 AND 2
t= s rut3........eqn 3

using eqn 1 and 3

s=+-1

hence B
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Prep Question...Need help 04 Jan 2010, 23:04
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How did you assume that O is origin?

xcusemeplz2009
OP=OQ=radius
OP=sqrt[(0-rut3)^2+(0-1)^2]=2-------------dis formula
radius=2---->OQ=2

OQ=sqrt[s^2+t^2]=4----->s^2+t^2=2....eqn 1

pythagoras theroem
OP^2+OQ^2=PQ^2

PQ^2= (s+RUT3)^2+(t-1)^2....DIS FORM
---->(s+RUT3)^2+(t-1)^2=4+4.........EQN 2
USING EQN 1 AND 2
t= s rut3........eqn 3

using eqn 1 and 3

s=+-1

hence B
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xcusemeplz2009
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Prep Question...Need help 04 Jan 2010, 23:58
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IN PS of GMAT for Geometry certain things can be assumed from the fig...
here X and Y denotes a coordinate system and Int of X and Y is origin
this is enough to assume

Had it been a DS , assuming the intersection as origin wud be wrng
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Prep Question...Need help 09 Jan 2010, 23:28
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xcusemeplz2009
OP=OQ=radius
OP=sqrt[(0-rut3)^2+(0-1)^2]=2-------------dis formula
radius=2---->OQ=2

OQ=sqrt[s^2+t^2]=4----->s^2+t^2=2....eqn 1

pythagoras theroem
OP^2+OQ^2=PQ^2

PQ^2= (s+RUT3)^2+(t-1)^2....DIS FORM
---->(s+RUT3)^2+(t-1)^2=4+4.........EQN 2
USING EQN 1 AND 2
t= s rut3........eqn 3

using eqn 1 and 3

s=+-1

hence B
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Mental road block. Please help.
OP=OQ b/c radii of circle
OP^2+OQ^2=PQ^2 *PQ=distance b/w P & Q

POQ isoceles right triangle...sides s,s, s sqrt 2
OP=distance b/w O & P OP^2= (-rt3-0)^2+(1-0)^2 OP^2=3+1=4 OP=2, OQ=2, PQ=2 sqrt 2
PQ^2=(s--rt3)^2+(t-1)^2
8=(s+rt3)^2 + (t-1)^2

Don't know where to go from here.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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