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intersect the y-axis

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abhi332 Really wants to Beat The GMAT!
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intersect the y-axis Post Wed Feb 24, 2010 6:01 am
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    Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

    (1) a^2+b^2>16
    (2) a=|b|+5

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    ldoolitt Really wants to Beat The GMAT! Default Avatar
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    Post Wed Feb 24, 2010 9:07 am
    abhi332 wrote:
    Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

    (1) a^2+b^2>16
    (2) a=|b|+5
    I believe the answer is (a) but I am not totally sure.

    What you really want to know is, given (1) or (2), does the above equation definitely have a solution at x=0 (the y intercept)

    For (1), let x=0. You have

    x^2 + y^2 - 2yb + b^2 = 16
    y^2 - 2yb = 16 - (a^2 + b^2)

    since a^2 + b^2 = 16 from (1), we know that the right hand side of that equation has to be negative. In other words

    y^2 - 2yb = some number less than zero
    y^2 - 2yb < 0

    y^2 < 2yb

    either

    y < 2b if y is positive
    y > 2b if y is negative

    There is a definite solution for these two possibilities.

    To check for (2), observe the 2 situations

    a = b+5 if b is positive
    a = 5-b if b is negative

    (b+5)^2 + (y-b)^2 = 16
    (y-b)^2 = 16 - (b+5)^2

    This obviously has a solution when b=-1 because that reduces to

    (y+1)^2 = 0, which y =-1

    But note that this has no solution when 16-(b+5)^2 < 0 because (y-b)^2 will always be a positive number. So this does NOT always yield the same answer. Thus (2) does not provide enough information.

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    Post Wed Feb 24, 2010 10:29 am
    abhi332 wrote:
    Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

    (1) a^2+b^2>16
    (2) a=|b|+5
    (x-a)^2 + (y-b)^2=16

    the curve is a a circle with a radius of 4 having center at (a,b)

    now for this to cut the y axis |a| <4

    1) Doesn't help - Insufficient

    2) a >5; sufficient the curve doesnt cut the y axis

    B

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    ldoolitt Really wants to Beat The GMAT! Default Avatar
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    Post Wed Feb 24, 2010 1:46 pm
    ajith wrote:
    abhi332 wrote:
    Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

    (1) a^2+b^2>16
    (2) a=|b|+5
    (x-a)^2 + (y-b)^2=16

    the curve is a a circle with a radius of 4 having center at (a,b)

    now for this to cut the y axis |a| <4

    1) Doesn't help - Insufficient

    2) a >5; sufficient the curve doesnt cut the y axis

    B
    That's a great way of looking at it. I didn't really see it that way.

    vineetbatra GMAT Destroyer!
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    Post Wed Feb 24, 2010 5:29 pm
    ajith wrote:
    abhi332 wrote:
    Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

    (1) a^2+b^2>16
    (2) a=|b|+5
    (x-a)^2 + (y-b)^2=16

    the curve is a a circle with a radius of 4 having center at (a,b)

    now for this to cut the y axis |a| <4

    1) Doesn't help - Insufficient

    2) a >5; sufficient the curve doesnt cut the y axis

    B
    Ajith, I am not sure if it is a stupid question, but how did you find that the curve is a circle. Also, after identifying its a circle how did you solve the rest of the problem.

    I will really appreciate if you can explain this in more detail. I am not very good in XY plane Q's.

    Vineet

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    Post Wed Feb 24, 2010 9:25 pm
    x*2 + y*2 = a*2 is the standard form of a circle.. (Read * as power of)


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    sivareddy Just gettin' started! Default Avatar
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    Post Wed Feb 24, 2010 9:26 pm
    and (x-a)*2 + (y-a)*2 = a*2 is a circle with centre at (a,b)

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    kstv GMAT Destroyer! Default Avatar
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    Post Thu Feb 25, 2010 1:58 am
    In this qs within the scope of GMAT ?

    ajith GMAT Titan
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    Post Thu Feb 25, 2010 2:18 am
    kstv wrote:
    In this qs within the scope of GMAT ?
    I do not think so

    I can see 2 ways to solve it
    1. Using geometry (involves equation of a circle etc etc)
    2. Using Quadratic Equations (Involves finding out whether quadratic equation has real roots etc..)

    Both of which are beyond the scope of GMAT.

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    Post Thu Feb 25, 2010 2:39 am
    ajith wrote:
    kstv wrote:
    In this qs within the scope of GMAT ?
    I do not think so

    I can see 2 ways to solve it
    1. Using geometry (involves equation of a circle etc etc)
    2. Using Quadratic Equations (Involves finding out whether quadratic equation has real roots etc..)

    Both of which are beyond the scope of GMAT.
    While I agree this isn't a "standard" GMAT problem, solving quadratics is certainly within the scope of the GMAT. I don't think its easy but it certainly COULD be asked.

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    Post Thu Feb 25, 2010 6:45 am
    @ Ajith

    acc to yr equations , u can assume that point a and point b will be less than 4 always. N this may be or may not is not the case here.


    they can be any point. Let say a , b are 8, 10 then even radius is 4 then ofcourse do not cut Y axis but let say

    point is 1 and 2 then circle will cut Y axis.



    plz put OA

    abhi332 Really wants to Beat The GMAT!
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    Post Thu Feb 25, 2010 6:48 am
    OA:B

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    ajith GMAT Titan
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    Post Thu Feb 25, 2010 6:53 am
    girish3131 wrote:
    @ Ajith

    acc to yr equations , u can assume that point a and point b will be less than 4 always. N this may be or may not is not the case here.

    plz put OA
    When did I make that assumption?

    I only said radius is 4 and the circle will cut y axis if the absolute value of a is less than 4. These are not assumptions, these are facts [There is a slight difference]

    a,b can be anything

    and 2) proves that a is greater than 5 - again no assumptions -

    I am not getting the "assumption" you talk about here

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    abhi332 Really wants to Beat The GMAT!
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    Post Thu Feb 25, 2010 6:58 am
    Ajit have solved this question beautifully Smile

    another way is

    curve will touch the y axis when x =0

    therefore, a^2 + (y-b)^2 = 16

    y^2 - 2yb + a^2 + b^2 -16 = 0

    b^2 - 4ac >= 0, checking the roots

    4b^2 - 4(1)(a^2 + b^2 -16) >=0
    a^2 <=16

    a^2-16<=0
    (a+4)(a-4)<=0

    a<= -4 or a<= 4

    |a| <= 4


    which is same what Ajit has found quickly Wink

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    amit.trivedi@ymail.com GMAT Destroyer!
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    Post Wed Mar 14, 2012 12:38 am
    What I understand from this question is that we need to know the distance between the center point of the circle and the y - axis i.e the y - intercept of the point.

    (2) a=|b|+5

    from this we do not care about the value for 'a' which is the x-intercept.

    So could any of the experts please guide whether i m correct or wrong.

    I am getting the value b as + or - 5. Hence the correct answer is B.

    Could any of you help...

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