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June 15th, 2009, 10:38 PM
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The Keeper of Grounds
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Q36. Math - Coordinate
Difficulty Level : Medium
Time to Solve : <60 secs
Please discuss your approach and also provide time you took to solve the question.
Q. A circle C is drawn with center as (0,0). A line L is tangent to the circle at point (-1, 1) , what is the slope of L?
A) 0
B) -1
C) 1
D) 1/2
E) None of those
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June 15th, 2009, 11:37 PM
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C ?
Took close to but less than 1 minute. Took a round about way . Looking for alternate methods from others .
Line L is perpendicular to the line joining the center of the circle (0,0) and point at which line L is a tangent. ( -1,1) ( definition of tangent.) Lets call this line R.
Slope of line R = (y2 -y1 ) / (x2-x1) = -1
Therefore slope of line L = ( Slope of line R ) * (-1 )
= 1
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June 16th, 2009, 01:35 AM
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C. yeah took less than 60 secs.
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June 17th, 2009, 12:19 AM
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the radius of the circle equals (1+1=2) - 2^(0,5)
therefore the equation of the circle x^2+y^2=2,
2x+2y*y'=0, so y'=-x/y
now let put our numbers: y'= -(-1/1)=1
therefore the slope of the tangent equals 1, C
or it's clear that the tangent touches the circle in the middle of circumference of the the circle in the left high quadrant (quart), therefore it has the angle 45, which is tag45=1
Last edited by clerk; June 17th, 2009 at 12:20 AM.
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June 17th, 2009, 06:04 AM
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C - 20 secs
Found the slope of the radius of the circle that touches the tangent point (-1,1). m = 1/-1 = -1. Since the line is perpendicular to the radius's slope the slope of the line is the opposite of the reciprocal. Therefore (-1/1)*-1 = 1.
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June 17th, 2009, 06:52 AM
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C
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June 17th, 2009, 10:57 AM
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Here is much easier way. The tangent to the circumference L will cross X axis at -1+x and also will cross Y axis at 1+y. Where -1+x=1+y. This tells us that we have an isosceles right triangle with angles 45,45,90. Thus tan45=1.
Dedicated to Clerk. Think easily
Quote:
Originally Posted by clerk
the radius of the circle equals (1+1=2) - 2^(0,5)
therefore the equation of the circle x^2+y^2=2,
2x+2y*y'=0, so y'=-x/y
now let put our numbers: y'= -(-1/1)=1
therefore the slope of the tangent equals 1, C
or it's clear that the tangent touches the circle in the middle of circumference of the the circle in the left high quadrant (quart), therefore it has the angle 45, which is tag45=1
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Last edited by TiTAN; June 17th, 2009 at 10:58 AM.
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June 18th, 2009, 07:34 AM
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GMAT Druid
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Quote:
Originally Posted by rkandell
C - 20 secs
Found the slope of the radius of the circle that touches the tangent point (-1,1). m = 1/-1 = -1. Since the line is perpendicular to the radius's slope the slope of the line is the opposite of the reciprocal. Therefore (-1/1)*-1 = 1.
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Good solutions everyone! Its amazing to see so many different solutions.
For those who dont know trigonometry , this solution by rkandell can be used as a quick method to solve the question.
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December 14th, 2009, 10:47 PM
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C
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December 19th, 2009, 11:33 PM
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C.
I didn't do any sort of paper work. If circle C touches (-1,1) it will also touch (1,1), (-1,1) and (-1,-1) having a center (0,0). Each joining line from (0,0) have a slope of 1 and slope at (-1,1) is parallel to the line from (-1,-1) to (1,1). Everything seems like diagonal of some squares which have slope of 1 each.
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