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19 Sep 2008, 15:34
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Attached is the question.
I don't fully understand what statement 1) tells us (not sure why the explanation about statement 1 is correct).
Thanks!
I don't fully understand what statement 1) tells us (not sure why the explanation about statement 1 is correct).
Thanks!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
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19 Sep 2008, 16:03
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foodstamp
I say that the answer is B.
First of all, when you look at the equation \((x-a)^2 + (y-b)^2 = 16\), the radius of the circle is 4. Also, the center of the circle is at (a,b). So we need to know the value of a and b (although a alone will be sufficient to help us answer this question) in order to figure out whether it's location is still within reach to the y-axis when taking its radius into consideration:
(1) \(a^2 + b^2 > 16\) ---> we can not determine the exact values for a and b because we can't tell whether they're positive or negative. Even if we do manage, the choices are endless
(2) \(a = |b| + 5\) ----> we know that |b| can be either a zero or positive. So when you rearrange this equation to:
\(a - 5 = |b|\), you know that the result must also be either a zero or positive. In order for this to be true, a must be at least a 5. When our value of a is at least a 5 and then we look at the center of the circle to be (a,b), the x-axis of 5 is longer than the length of the radius of 4, therefore the answer is no: the circle does not intersect the y-axis.
Answer is B.
19 Sep 2008, 23:13
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a^2+b^2 > 16.
a, b can be anything. if a=0 , b =5, yes it intersects the y axis.
if a= 2000, b=4000(some random value), it does not
stat 2)when b=0 a = 5. so the min point is 5,0. and here the circle does not intersect.
and b can be +ve or -ve. but x is always +ve.
so the point lies in the 1st or 4th quadrant and the point is 4 units away from the y axis. suff.
ans is B.
a, b can be anything. if a=0 , b =5, yes it intersects the y axis.
if a= 2000, b=4000(some random value), it does not
stat 2)when b=0 a = 5. so the min point is 5,0. and here the circle does not intersect.
and b can be +ve or -ve. but x is always +ve.
so the point lies in the 1st or 4th quadrant and the point is 4 units away from the y axis. suff.
ans is B.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
