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In the rectangular coordinate system, points (4, 0) and [#permalink]
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11 Nov 2009, 13:13
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Re: Maximum value of the radius [#permalink]
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Bunuel wrote: In the rectangular coordinate system, points (4, 0) and (– 4, 0) both lie on circle C. What is the maximum possible value of the radius of C ?
(A) 2 (B) 4 (C) 8 (D) 16 (E) None of the above I'm getting E It can be B, but the points mentioned can be a chord and that would make the radius larger. I'm getting other calculations but none are available or can't be determined.



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Re: Maximum value of the radius [#permalink]
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11 Nov 2009, 16:04
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The answer is E.
It takes 3 distinct points to define a circle. Only 2 are given here.
The two points essentially identify a single chord of the circle C. Since no other information is provided, however, the radius of the circle can essentially be anything. All this information tell us is that the radius is greater than 4. It does not give us an upper limit.



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Re: Maximum value of the radius [#permalink]
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11 Nov 2009, 17:48
Agree with E.
Another way to look at it is that the two points, lets call them A and B, are equidistant to the centre of the circle, lets call that O. i.e. OA = OB Hence the centre will lie on the Y axis (anywhere where x = 0). So not enough information to determine.
Yet another way to look at it is: Radius^2 = (Difference of X of O to A)^2 + (Difference of Y of O to A)^2 From the question stem we know that A is (4,0). Using the above logic we also know that the centre lies on x=0. Using B would yield the same result as we are after distance it will always end up being positive anyway. This formula reduces to (40)^2 + (y0)^2 = R^2 Depending on the value of y, the length of the radius will keep growing.



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Re: Maximum value of the radius [#permalink]
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11 Nov 2009, 20:42
Yep E , we need to know the origin to determine the radius and from the above information we cannot determine the origin.
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Re: Maximum value of the radius [#permalink]
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Re: Maximum value of the radius [#permalink]
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19 Sep 2010, 01:43
The question is more like a DS question. Rephrase it and you will get "If two points given are enough to define the maximum possible radius of the circle?" The answer is no, cause the radius could be as low as 4 if the points are at the maximum distance from the center and the line between them is the diameter or the radius could be infinitely large if the line between the points is the chord.



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Re: Maximum value of the radius [#permalink]
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04 Nov 2010, 15:28
You made me think a lot..But I arrived at (E) since the two points can be a chord too... IMO is (E) for me too
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Re: Maximum value of the radius [#permalink]
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29 Nov 2010, 02:06
Bunuel wrote: The OA is E.
The only thing we can conclude from the question that center lies on the Yaxis. But it could be ANY point on it, hence we can not determine maximum value of r. Can we also conclude that the points (4,0) and (4,0) lie in first and 2nd quadrant so with that we cannot calculate the distance between two points ( which will be radius of circle ) ; because in order to calculate distance we need points in opposite direction. So if the points were in Ist and 3rd quadrant we could have calculated the distance



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Re: Maximum value of the radius [#permalink]
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29 Nov 2010, 02:22
rite2deepti wrote: Bunuel wrote: The OA is E.
The only thing we can conclude from the question that center lies on the Yaxis. But it could be ANY point on it, hence we can not determine maximum value of r. Can we also conclude that the points (4,0) and (4,0) lie in first and 2nd quadrant so with that we cannot calculate the distance between two points ( which will be radius of circle ) ; because in order to calculate distance we need points in opposite direction. So if the points were in Ist and 3rd quadrant we could have calculated the distance I think you are a little bit confused here. You CAN calculate the distance between any two points with given coordinates on a plane (no matter in which quadrants they are). For example the distance between two points (4,0) and (4,0) is simply 8. Generally the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). Next, the distance between (4,0) and (4,0) won't necessarily be the DIAMETER of a circle. The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the yaxis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all. For more check Coordinate Geometry chapter of Math Book: mathcoordinategeometry87652.htmlHope it helps.
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Re: Maximum value of the radius [#permalink]
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06 Jan 2011, 16:33
if we join the line connecting the the points (4,0) and (4,0) to the center of the circle say (0,y), radius will be maximum at the point where the area formed by the above triangle is min. The area will be 0 if the height is 0, which means the center is in the line connecting two pnts (4,0) and (4,0). Isn't?



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Re: Maximum value of the radius [#permalink]
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06 Jan 2011, 18:17
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praveenvino wrote: if we join the line connecting the the points (4,0) and (4,0) to the center of the circle say (0,y), radius will be maximum at the point where the area formed by the above triangle is min. The area will be 0 if the height is 0, which means the center is in the line connecting two pnts (4,0) and (4,0). Isn't? The red part is not correct. Again: The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the yaxis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all. Check 2 possible circles: Circle with min radius of 4 (equation x^2+y^2=4^2): Attachment:
radius 4.gif [ 3.22 KiB  Viewed 19564 times ]
Circle with radius of 5 (equation x^2+(y3)^2=5^2): Attachment:
radius 5.gif [ 2.78 KiB  Viewed 19564 times ]
Generally circle passing through the points (4, 0) and (– 4, 0) will have an equation \(x^2+(ya)^2=4^2+a^2\) and will have a radius of \(r=\sqrt{4^2+a^2}\). As you can see min radius will be for \(a=0\), so \(r_{min}=4\) and max radius is not limited at all (as \(a\) can go to +infinity as well to infinity). For more check Coordinate Geometry chapter of Math Book: mathcoordinategeometry87652.html Hope it helps.
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Re: Maximum value of the radius [#permalink]
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27 Apr 2011, 00:42
[quote="Mongolia2HBS"]Only 2 points are given, so E.[/quo
suppose the question says that what is the maximum possible radius of the among the given choices then we can try out the highest choices and foind out whether the the given choice satisfies the pythagoras eqn. so that way D is best.



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Re: Maximum value of the radius [#permalink]
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25 Apr 2012, 11:38
hm, maybe it is not an elegant method, but still... we have two points  (4, 0) and (– 4, 0) . Also we know the formula (xa)^2+(yb)^2=r^2 (4a)^2+(0b)^2=(4a)^2+(0b)^2 (4a)^2=(4a)^2 a=0 since we cant find b, we have no clue about the position of the center. So, no chances to find the maximum possible value of the radius hope my logic is ok. let me know, if I am wrong
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Re: Maximum value of the radius [#permalink]
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25 Apr 2012, 12:31
LalaB wrote: hm, maybe it is not an elegant method, but still...
we have two points  (4, 0) and (– 4, 0) . Also we know the formula (xa)^2+(yb)^2=r^2
(4a)^2+(0b)^2=(4a)^2+(0b)^2
(4a)^2=(4a)^2 a=0
since we cant find b, we have no clue about the position of the center. So, no chances to find the maximum possible value of the radius
hope my logic is ok. let me know, if I am wrong Here is how can you solve this question with that approach: intherectangularcoordinatesystempoints40and86703.html#p847714
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Re: In the rectangular coordinate system, points (4, 0) and [#permalink]
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27 Jun 2013, 23:47



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Re: In the rectangular coordinate system, points (4, 0) and [#permalink]
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28 Jun 2013, 04:57
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Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREThe only thing we know is that Points A(4,0) and B(4,0) are on circumference. But that does not necessarily mean that they are opposite ends of diameter. If they are opposite ends of diameter, Radius will be 4, but if they are opposite ends of circle's smallest chord then Radius would be far more greater then the values mentioned in Options. Hence Choice E is correct.
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Re: Maximum value of the radius [#permalink]
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25 Oct 2013, 08:45
Bunuel wrote: rite2deepti wrote: Bunuel wrote: The OA is E.
The only thing we can conclude from the question that center lies on the Yaxis. But it could be ANY point on it, hence we can not determine maximum value of r. Can we also conclude that the points (4,0) and (4,0) lie in first and 2nd quadrant so with that we cannot calculate the distance between two points ( which will be radius of circle ) ; because in order to calculate distance we need points in opposite direction. So if the points were in Ist and 3rd quadrant we could have calculated the distance I think you are a little bit confused here. You CAN calculate the distance between any two points with given coordinates on a plane (no matter in which quadrants they are). For example the distance between two points (4,0) and (4,0) is simply 8. Generally the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). Next, the distance between (4,0) and (4,0) won't necessarily be the DIAMETER of a circle. The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the yaxis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all. For more check Coordinate Geometry chapter of Math Book: mathcoordinategeometry87652.htmlHope it helps. Can you please explain on what basis it is concluded that centre lies on Y axis. Secondly which part in Gmat math book the concept is given, I have gone through the book but I didn't find it. Third can you please explain what is the concept which clarifies minimum point to make a circle & maximum radius concept. Please Help



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Re: Maximum value of the radius [#permalink]
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25 Oct 2013, 08:54
anu1706 wrote: Bunuel wrote: rite2deepti wrote: Can we also conclude that the points (4,0) and (4,0) lie in first and 2nd quadrant so with that we cannot calculate the distance between two points ( which will be radius of circle ) ; because in order to calculate distance we need points in opposite direction. So if the points were in Ist and 3rd quadrant we could have calculated the distance I think you are a little bit confused here. You CAN calculate the distance between any two points with given coordinates on a plane (no matter in which quadrants they are). For example the distance between two points (4,0) and (4,0) is simply 8. Generally the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). Next, the distance between (4,0) and (4,0) won't necessarily be the DIAMETER of a circle. The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the yaxis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all. For more check Coordinate Geometry chapter of Math Book: mathcoordinategeometry87652.htmlHope it helps. Can you please explain on what basis it is concluded that centre lies on Y axis. Secondly which part in Gmat math book the concept is given, I have gone through the book but I didn't find it. Third can you please explain what is the concept which clarifies minimum point to make a circle & maximum radius concept. Please Help Please check here: intherectangularcoordinatesystempoints40and86703.html#p847714
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Re: In the rectangular coordinate system, points (4, 0) and [#permalink]
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