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 ?
A. 1.4
B. 2^1/2
C. 1.7
D 3^1/2
E. 2.0
Please explain the answer.
OA is A
Shortest Distance between a Circle and a Line
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- ssmiles08
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I sort of got the answer...but my method was somewhat crude...maybe some one has a better answer for you.
First I figured out the x and y intercepts of the line. (4,0) and (-3,0). this told me right away that there is some sort of a 3-4-5 triangle involved with it.
you know the radius of the circle is 1 with origin at (0,0).
when you draw it out in paper, you would have a right triangle in the 4th quadrant with 1/4 of the circle coming in the way.
The shortest distance would be the line perpendicular to the line Y=3/4X-3 and the circle.
so what I did was draw out a rectangle at the 4th quadrant (basically two 3-4-5 triangles) (0,0)(4,0)(-3,0)(4,-3)
Then I drew two diagonals with the length of 5 within the rectangle. The distance from the origin(0,0) to the midpoint of the rectangle is approximately 2.5.
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
Like I said, my method was crude and I am not even sure if it right...maybe someone has a better explanation.
First I figured out the x and y intercepts of the line. (4,0) and (-3,0). this told me right away that there is some sort of a 3-4-5 triangle involved with it.
you know the radius of the circle is 1 with origin at (0,0).
when you draw it out in paper, you would have a right triangle in the 4th quadrant with 1/4 of the circle coming in the way.
The shortest distance would be the line perpendicular to the line Y=3/4X-3 and the circle.
so what I did was draw out a rectangle at the 4th quadrant (basically two 3-4-5 triangles) (0,0)(4,0)(-3,0)(4,-3)
Then I drew two diagonals with the length of 5 within the rectangle. The distance from the origin(0,0) to the midpoint of the rectangle is approximately 2.5.
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
Like I said, my method was crude and I am not even sure if it right...maybe someone has a better explanation.
perpendicular Distance(shortest distance) of a point (x1,y1) from a line ax+by+c = 0 is given by
d = |(a*x1)+(b*y1)+c|/(a^2+b^2)^1/2
Let us calculate the distance of the line from the center of the circle (0,0)
a = -3/4
b=1
c=3
x1=0
y1=0
therefore d= 12/5
distance from a point on the circle would be 12/5-1 = 1.4
hope this helps!!
d = |(a*x1)+(b*y1)+c|/(a^2+b^2)^1/2
Let us calculate the distance of the line from the center of the circle (0,0)
a = -3/4
b=1
c=3
x1=0
y1=0
therefore d= 12/5
distance from a point on the circle would be 12/5-1 = 1.4
hope this helps!!
Last edited by Claret on Mon Jun 29, 2009 8:08 pm, edited 1 time in total.
one thing to note here in this question is two answer choices are pretty closessmiles08 wrote:
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
A) 1.4 and B) (2) ^1/2 ~ 1.414
if we bring down the calculation to ~1.5 , either A or B could be the answer..
- ssmiles08
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Yes you are right. Thank you.Claret wrote:one thing to note here in this question is two answer choices are pretty closessmiles08 wrote:
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
A) 1.4 and B) (2) ^1/2 ~ 1.414
if we bring down the calculation to ~1.5 , either A or B could be the answer..
sanju, Ian, Kowinsky, please comment, found it very difficult for gmat
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The answer is indeed 1.4 but I think it needs to be solved differently.
I think everybody has figured out the vertices of traingle OAB that is O (0,0), A(0,-3) and B(4,0). Now draw a perpendicular line from O to AB (Sorry I have not drawn fig). Say it touches OB at D.
Now you can find out OD= OB*Sin <OAB
= 4*3/5
= 12/5
We know that a tangent on any point on the circle is perpendicular to the radius.
Therefore, the shortest distance from any point on circle to the line AB= OD- Radius of the circle
i.e. 12/5-1= 1.4
I think everybody has figured out the vertices of traingle OAB that is O (0,0), A(0,-3) and B(4,0). Now draw a perpendicular line from O to AB (Sorry I have not drawn fig). Say it touches OB at D.
Now you can find out OD= OB*Sin <OAB
= 4*3/5
= 12/5
We know that a tangent on any point on the circle is perpendicular to the radius.
Therefore, the shortest distance from any point on circle to the line AB= OD- Radius of the circle
i.e. 12/5-1= 1.4