This is a tough problem. I think most people are going to work on it for 2 min on test day and then make an educated guess.

The circle has 0,0 as the origin and a radius of 1. The line has y-intercept of -3 and runs up and to the right with a slope of +3/4. So it also has coordinates of (0,0), (0,-3) and (0,4). This means we have a 3-4-5 triangle.

The question asks for the least possible distance (a.k.a. the closes point) between the circle and the line given by the formula. The closest point of the circle to the line can be found by creating a second line. The slope of this line must be inverted and negated (slope of 3/4, becomes -4/3) and the y-intercept will be 0, meaning the line goes through the origin. Where this newly created line passes through the circle will be the closest point to the line. Now we just have to figure out the distance between these two points.

I first started off by drawing a picture. Sorry my photoshop isn't working or I'd upload a picture. But first draw the circle with radius 1 around the origin. I made 1 point at 0,-3 for the line's y-intercept and one dot at 4,0 for the x-intercept. I then drew a line as a hypotnuse of a triangle created by the x & y axis and the line. The hypotnuse I labeled c, the line from 0,0 to 0,-3 i labeled a and the line from 0,0 to 4,0 i labeled b. So we have triangle with sides a, b, and c.

Now the prependicular line created by y = -4/3x (inverted and negated slope of the line given) will divide this hypotnuse. I labeled the distance from the origin to the point where the original line and perpendicular cross as x. I also labeled the distance from (0,-3) to the perpendicular as d and the point from the perpendicular to (4,0) as e.

Because we know side a = 3, side b = 4, this means side c = 5

So we have \(a^2 + b^2 = c^2\), \(d^2 + x^2 = a^2\), and \(e^2 + x^2 = b^2\) do some substitutions.

d^2 + x^2 = 9 & e^2 + x^2 = 16.

Because side c = d + e, we know that c = 5, so 5 = d + e. This can also be written d = 5 - e and e = 5 - d. These will become important later.

I've found it easier to combine 2 equations where possible. This can be done with both equations = the same thing, like x^2.

So our perpendicular created 2 smaller triangles within the larger one. Trianle ADX and EBX. Let's represent these by:

d^2 + x^2 = a^2 & e^2 + x^2 = b^2 Sub in the values we know for a and b

d^2 + x^2 = 9 & e^2 + x^2 = 16

x^2 = 9 - d^2 and x^2 = 16 - e^2 Combine these;

x^2 + x^2 = 9 - d^2 + 16 - e^2; we have the variable d in there that we need to find a way to sub in to allow us to solve

d = 5 - e (from above) so d^2 = (5 - e)^2

2x^2 = 9 - (5-e)^2 + 16 - e^2; do FOIL on (5-e)^2

2x^2 = 9 - (25 - 10e + e^2) + 16 - e^2; now get rid of those ( ) by factoring the - into the terms

2x^2 = 9 -25 + 10e - e^2 + 16 - e^2. We see that we have e and e^2. Now what? well, we have values we can sub in for e^2 that include an x as a variable. This is good because we don't want to substitute in and keep changeing variables and never get done subbing in variables. So what does e^2 equal in terms of x? e^2 = 16 - x^2 so now sub this in

2x^2 = 9 - 25 + 10e - (16 - x^2) + 16 - (16 - x^2); The parenthesis are important because we're subbing and we don't want this to throw our values off. If the term subbed into the problem is + but () are not used, then the second term will not be correct in terms of + or -. Do some cleaning up to factor in and combine terms were possible.

2x^2 = -16 + 10e - 16 + x^2 + 16 - 16 + x^2; do some housecleaning and combine terms

2x^2 = 10e - 32 + 2x^2

32 = 10e

e = 3.2

remember that e is the distance from (4,0) to the perpendicular. This can now be used to find the value of x because e^2+x^2 = b^

(3.2)^2 + x^2 = 16

3.2 * 3.2 = 10.24

10.24 + x^2 = 16

x^2 = 5.76

x = 2.4

Now we're not done (for reason #1, 2.4 doesn't match any answer!) But this is the distance from the perpendicular to the origin. But we're looking for the closest point in the circle to the line. What was the radius of the circle? 1 so we need to take 2.4 - 1 to find the distance of the point on the actual circle to the line.

Answer A.

There may be faster ways to answer this on test day, but if you get something like this, take a look to see if there is anything obvious, but then guess and move on. You're better off not wasting your time.

ritula wrote:

What is the least possible distance between a point on the circle x^2+y^2=1 and a point on the line y=3/4x-3 ?

1.4

sqrt(2)

1.7

sqrt(3)

2

_________________

------------------------------------

J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings