OG - prep test
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Source: Beat The GMAT — Problem Solving |
- jayhawk2001
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Equation of line connecting (-sqrt3, 1) to origin is
y -1 = -1/sqrt-3 (x + sqrt-3)
y = -x/sqrt-3
Line perpendicular to the above line will have slope of +sqrt-3
So, y = sqrt-3 * x
This line touches origin and point (s,t), so
t = sqrt-3 * s
t^2 = 3*s^2
We also know that radius of circle = 2 (i.e. sqrt [ (sqrt-3)^2 + 1 ])
So, s^2 + t^2 = 4
s^2 + 3*s^2 = 4
s^2 = 1
s = 1
y -1 = -1/sqrt-3 (x + sqrt-3)
y = -x/sqrt-3
Line perpendicular to the above line will have slope of +sqrt-3
So, y = sqrt-3 * x
This line touches origin and point (s,t), so
t = sqrt-3 * s
t^2 = 3*s^2
We also know that radius of circle = 2 (i.e. sqrt [ (sqrt-3)^2 + 1 ])
So, s^2 + t^2 = 4
s^2 + 3*s^2 = 4
s^2 = 1
s = 1












