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29 Oct 2007, 10:13
Kudos
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What is the least possible distance between a point on circle x^2 + y^2 = 1 and a point on the line y=(3/4)X -3?
1.4
sqrt(2)
1.7
sqrt(3)
2
What is the formula?
1.4
sqrt(2)
1.7
sqrt(3)
2
What is the formula?
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29 Oct 2007, 11:20
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i do not think there's any direct formula involved here...i know a method to solve this problem..the answer is (a) 1.4...
I do not know if there is an alternate method to solve this but this is the one i thought of.
you should probably be drawing a diagram for this one.....basically, the given equation of the circle is that of a circle with a radius of 1 and center at the origin(0,0)(call this point O)(draw this, seriously!!).
The given line's equation give you two points on the X and the Y axis. The two points will be (4,0) on the X axis(call this one point A) and (0,-3) on the Y axis(call this one point B).
Now the least possible distance would be a perpendicular from the circle to the line...in other words, a perpendicular from the center of the circle(the origin) to the given line......this perpendicular line would have a slope of (-4/3) and an equation of y=(-4X/3).....using the equations of the given line and the perpendicular, we get an intersection between the two lines at point C(1.44,1.92).
If you have drawn a diagram for this, you will see that there is a right angled triangle formed between the points of the given line intersecting the X and the Y axis and the origin. ie. triangle OAB. According to the rules of geometry (OC)^2=(AC)*(BC).
i.e..(OC)^2=5.76
therefore OC=2.4 units.
Do not forget to subtract the radius from this length.
Therefore,least possible distance is 2.4-1=1.4.
Hope that helps.Good luck.
I do not know if there is an alternate method to solve this but this is the one i thought of.
you should probably be drawing a diagram for this one.....basically, the given equation of the circle is that of a circle with a radius of 1 and center at the origin(0,0)(call this point O)(draw this, seriously!!).
The given line's equation give you two points on the X and the Y axis. The two points will be (4,0) on the X axis(call this one point A) and (0,-3) on the Y axis(call this one point B).
Now the least possible distance would be a perpendicular from the circle to the line...in other words, a perpendicular from the center of the circle(the origin) to the given line......this perpendicular line would have a slope of (-4/3) and an equation of y=(-4X/3).....using the equations of the given line and the perpendicular, we get an intersection between the two lines at point C(1.44,1.92).
If you have drawn a diagram for this, you will see that there is a right angled triangle formed between the points of the given line intersecting the X and the Y axis and the origin. ie. triangle OAB. According to the rules of geometry (OC)^2=(AC)*(BC).
i.e..(OC)^2=5.76
therefore OC=2.4 units.
Do not forget to subtract the radius from this length.
Therefore,least possible distance is 2.4-1=1.4.
Hope that helps.Good luck.
31 Oct 2007, 06:57
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sorry for the typo....point C should be (1.44,-1.92)...
point C is derived by simultaneously solving the equation of the given line(y=(3/4)X -3) with the equation of the line perpendicular to it(y=(-4X/3))...
point C is derived by simultaneously solving the equation of the given line(y=(3/4)X -3) with the equation of the line perpendicular to it(y=(-4X/3))...
