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    The Keeper of Grounds Hagrid's Avatar
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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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    Just starting GMAT Prep guys.from.fresno's Avatar
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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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    Just starting GMAT Prep marcus fenix's Avatar
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    C. yeah took less than 60 secs.

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    Ready to Crack GMAT clerk's Avatar
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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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    Serious about GMAT Prep rkandell's Avatar
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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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    Just starting GMAT Prep pssohal's Avatar
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    C

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    Just starting GMAT Prep TiTAN's Avatar
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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 View Post
    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 TiTAN; June 17th, 2009 at 10:58 AM.

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    GMAT Druid Guardian's Avatar
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    Quote Originally Posted by rkandell View Post
    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.
    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.
    ~|Guardian|

    I am an instructor on ScoreChase GMAT Crash Course . I also provide GMAT Private Tutoring sessions.

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    Really wants to crack GMAT ! GMAT_Dreamer's Avatar
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    Just starting GMAT Prep Jilany's Avatar
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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.
    Nurul Hai

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