killer........

g4gmat 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



answer a

The circle described has a radius of 1, and is centred on the origin (0,0).

The line described intercepts the y-axis at (0,-3), and the x-axis at (4,0). Therefore the line, plus the x- and y-axes define a right-angled triangle with sides of length 3, 4, and 5.

The height of the triangle (with respect to the long edge) is a line segment from the origin to the line y = 3/4*x - 3. Part of that height is within the radius of the circle, and the rest is the minimum distance from the circle to the line.

The part within the radius must be 1 unit long. The easiest way to calculate the height of the triangle that I can see is to use the area formula, area = 0.5 * b * h

Firstly area = 0.5 * 3 * 4 = 6

And then, keeping area constant, but using the long edge as the base:

area = 0.5 * b * h
6 = 0.5 * 5 * h
h = 12 / 5 = 2.4

Remember to subtract the 1 that is within the radius of the circle, giving an answer of 1.4

To "cjb" this is really confusing...can u explain it with the help of a diagram...also i didn't understand a part in the quest that says x^2 +y^2=1....why did u take 1 as the radius of a circle??

Check this for the equation of a circle:

https://en.wikipedia.org/wiki/Circle#Equation

Another way to see that this is a circle is to think about Pythagoras's Theorem. The points that satisfy the equation x^2 + y^2 = 1 are all the points that are a distance of 1 from the origin. The distance of any point from the origin is the square root of the sum of the squares of its x- and y-ordinates.

I don't have a diagram tool - what do other people use?

cjb..u can attach file with the diagram while posting...neways no need..it is much clear now..circle equation was something new to me ...got the basic from that wikipedia...it has got some insane properties, i hope most of them r not required in gmat..is it?

g4gmat wrote:cjb..u can attach file with the diagram while posting...neways no need..it is much clear now..circle equation was something new to me ...got the basic from that wikipedia...it has got some insane properties, i hope most of them r not required in gmat..is it?

No problems attaching a file - I just don't know how to produce it to begin with!

No idea what properties of circles can be tested in the GMAT.

g4gmat 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
answer a

distance between any point and a straight line will be the least when a perpendicular to the st line passes through that point.
the given circle has origin (0,0) as the centre and radius =1
after simplifying the given eq. of the st. line we get
3x-4y-12=0
perpendicular distance of this st line from the origin=
|3*0-4*0-12|/sqrt(3^2+4^2)
=12/5 (i've used the formula of distance of a st. line from a point)
therefore the least distance= (12/5)-1=7/5=1.4