x2suresh wrote:

durgesh79 wrote:

GMAT TIGER 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 = 3/4*x - 3?

A) 1.4

B) sqrt (2)

C) 1.7

D) sqrt (3)

E) 2.0

shortest distance from (0,0) to the line - radius of the circle

=12/5 - 1

= 7/5

= 1.4

Can you elaborate

The line containing the shortest path b/w a circle and line X is the line that is perpindicular to line X and passes through the center of the circle. To calculate the distance b/w the circle and line X, you take the distance from the center of the circle to line X and subtract out the distance from the center of the circle to the edge of the circle (i.e., the radius).

I did it a long way. Durgesh, I don't know if you have a quicker way. This took me more than 2 minutes.

The line, with x and y axes create a 3-4-5 triangle. The non-leg base of this right triangle extends from (0,0) and intersects the hypotenuse / line at a right angle. This base creates 2 new right triangles with hypotenuses created by the x- and y-axis, with lengths 4 and 3, respectively.

Let a = the length of the upper half of the hypotenuse of length 5 created by the intersection of the line and the non-leg base

Let b = the length of the non-leg base

a^2 + b^2 = 4^2 = 16

(5-a)^2 + b^2 = 3^2 = 9

25 - 10a + a^2 + b^2 = 9

25 - 10a + 16 = 9

10a = 32

a = 16/5

(16/5)^2 + b^2 = 16

b^2 = 16 - 16^2/25 = 16*(1-16/25) = 16*9/25

b = sqrt(16*9/25) = 4*3/5 = 12/5

Distance = 12/5 - 1 = 7/5