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geometry circle

This topic has 1 expert reply and 2 member replies
chamisool Junior | Next Rank: 30 Posts Default Avatar
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geometry circle

Post Thu Oct 27, 2011 8:19 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])



    The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?


    (A) k
    (B) k/(sqrt 2)
    (C) k/(sqrt 3)
    (D) k/2
    (E) k/3

    Can anyone help me with this question? I understand the answer but the answer I originally got was not in the answer choices.

    What I did was I extended k past the center of the circle, so I form the circle inscribed in a square.

    As k is double the length, I made it so its 2k as the hypotenuse. The ratio formula for an equilateral right triangle is x : x : x*sqrt 2
    So I did x*sqrt 2 = 2k
    Isolate x and you get x = 2k/(sqrt 2)

    So the legs of the triangle is 2k/(sqrt 2). I simplified it by multiplying both numerator and denominator by sqrt 2, so I got
    X = (2k*sqrt 2)/2 which becomes k*sqrt 2. Since the leg is the diameter, I divided it by 2 and my final answer was (k*sqrt 2)/2 but the answer choice isn’t there.

    Is it wrong to simplify 2k/(sqrt 2)? Because if I get 2k/sqrt 2 divided by 2, then I get the correct answer. I guess my question is when do you leave the sqrt denominator alone and when do you multiply both numerator and denominator by the sqrt?

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    Amiable Scholar Senior | Next Rank: 100 Posts
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    Post Thu Oct 27, 2011 8:27 am
    the side of your square is not the radius but equal to diameter of the circle. Smile
    so you need to divide it by 2
    so radius is
    (2k/sqrt(2))/2
    =k/sqrt(2)

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    Anurag@Gurome GMAT Instructor
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    Post Thu Oct 27, 2011 8:28 am
    chamisool wrote:
    ... my final answer was (k*sqrt 2)/2 but the answer choice isn’t there.
    2 = (√2)*(√2)

    Hence, √2/2 = 1/√2
    Hence, k√2/2 = k/√2 --> Option B

    As far as solving the problem is concerned, there is no need to extend k and bring complications. Just draw a perpendicular from C on the x-axis. Then you will get an isosceles right angled triangle with hypotenuse of length k.

    Hence, length of the other sides = length of radius = k/√2

    The correct answer is B.

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    chamisool Junior | Next Rank: 30 Posts Default Avatar
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    Post Thu Oct 27, 2011 8:55 am
    Amiable Scholar wrote:
    the side of your square is not the radius but equal to diameter of the circle. Smile
    so you need to divide it by 2
    so radius is
    (2k/sqrt(2))/2
    =k/sqrt(2)
    Yeah, that is what I did. But before I divided the diameter by 2 I multiplied the denominator and numerator by sqrt 2 to simply my answer.

    So I got 2k/(sqrt 2) *(sqrt 2/sqrt 2)

    So I got (2k*sqrt 2)/2 = k*sqrt 2

    So I got diameter of k*sqrt 2 and then divided that by 2 to find the radius. (k*sqrt 2)/2

    I guess I'm confusing myself cause I noticed based Anurag's response, if I multiply the numerator and denominator again, the answer becomes k/sqrt 2.

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