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What is the least possible distance between a point on the

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What is the least possible distance between a point on the [#permalink] New post 07 Jun 2007, 20:35
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A
B
C
D
E

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What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3x/4 - 3?
A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
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 [#permalink] New post 07 Jun 2007, 22:04
I got A
First the least distance is a perpendicular from the centre of the circle to a point on the line minus redius.
The point of intersection of the line with the x-axis is 4, with y-exis - -3
So we have a right triangle (3,4,5).
Then we draw the height from centre of the circle (this is also point (0,0)) and it will be the least distance between (0,0) and a poit on the line.
We find the height and then deduct radius 1.
The most time-consuming part for me was to calculate the height.
Is there any shortcut to calculate the height of a right triangle if we know all its sides?
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 [#permalink] New post 08 Jun 2007, 21:00
Give me A! :-D
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 [#permalink] New post 09 Jun 2007, 09:22
Answer is 'A'.
But it took me long to solve this.

Solve the equations : y = 3x/4 -3 and y = -4/3 x
y = -4/3x is the equation of the line passing through origin and perpendicular to the given line.

Solving them, the foot of the perpendicular drawn from the origin to the line = (36/25, -48/25).

Calculate the distance between origin and (36/25, -48/25) and subtract 1.
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 [#permalink] New post 09 Jun 2007, 10:04
I got A.

Plotting the points, you get a circle with r=1, and a 3,4,5 triangle. The shortest point between them would be a line drawn 45 degrees from the origin... when plotting this line, you can deduct that it must be slightly smaller than 2.5. Calculating you get 2 2/5 or 2.4... 2.4-1 gives you 1.4

On the actual GMAT, I think they would space these answers out a bit more to make 1 clear answer... distinguishing between sqrt2, and 1.4 seems a bit trivial.
  [#permalink] 09 Jun 2007, 10:04
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