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Is the radius of the circle greater than 3? [#permalink]
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19 Nov 2010, 04:33
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Is the radius of the circle greater than 3? (1) (2,4) and (5,10) lie on the circle. (2) (2,4) and (4,1) lie on the circle. My understanding is that the question asks us whether the radius of the circle greater than 3 or not? Thus if we are able to get a Yes or No from the option, it should suffice.
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Re: A DS problem [#permalink]
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19 Nov 2010, 04:55
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gdk800 wrote: Hi All,
Please help me understand the following problem.
Question: Is the radius of the circle greater than 3? 1) (2,4) and (5,10) lie on the circle. 2) (2,4) and (4,1) lie on the circle.
My understanding is that the question asks us whether the radius of the circle greater than 3 or not? Thus if we are able to get a Yes or No from the option, it should suffice. Yes, your understanding is correct. Maximum distance between two points on a circle is the length of its diameter (so when these points are the endpoints of a diameter), so the max distance = diameter = 2 radii. So for example if we are told that the distance between the two points on a circle is 10cm then we can be sure that the diameter is more than or equal to 10 (or the radius is more than or equal to 10/2=5). Next, the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). Now, back to the original question: Is the radius of the circle greater than 3?1) (2,4) and (5,10) lie on the circle > the distance between these points is \(d=\sqrt{(25)^2+(410)^2}=\sqrt{45}>6\), so the diameter of this circle must be more than 6, thus the radius must be more than 3 > answer to the question is YES. Sufficient. 2) (2,4) and (4,1) lie on the circle > the distance between these points is \(d=\sqrt{(24)^2+(41)^2}=\sqrt{13}<6\), so the diameter of this circle may or may not be more than 6. Not sufficient. Answer: A.
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Re: A DS problem [#permalink]
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19 Nov 2010, 08:47
Hi Much appreciate you posting back, here is my confusion on this.
In the second case, if Diameter < 6 than the radius (Diameter /2) must also be < 3 and thus is answers our question that the radius is not greater than 3. Hence i marked D instead of A...
Please clarify this doubt.



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Re: A DS problem [#permalink]
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19 Nov 2010, 08:58
gdk800 wrote: Hi Much appreciate you posting back, here is my confusion on this.
In the second case, if Diameter < 6 than the radius (Diameter /2) must also be < 3 and thus is answers our question that the radius is not greater than 3. Hence i marked D instead of A...
Please clarify this doubt. First of all: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. Which means that you can not have YES answer for statement (1) and NO answer for statement (2). Next, the fact that the distance between two points is \(\sqrt{13}\) just means that \(diameter\geq{\sqrt{13}}\approx{3.6}\), diameter can not be less than this value (radius can not be less than ~1.8) but it can be more than it. The same for (1): the distance between two pints is \(\sqrt{45}\approx{6.7}\), so diameter can not be less than this value (so radius can not be less than ~3.35) but it can be more than it. Hope it's clear.
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Re: A DS problem [#permalink]
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19 Nov 2010, 09:42
gdk800 wrote: Hi All,
Please help me understand the following problem.
Question: Is the radius of the circle greater than 3? 1) (2,4) and (5,10) lie on the circle. 2) (2,4) and (4,1) lie on the circle.
My understanding is that the question asks us whether the radius of the circle greater than 3 or not? Thus if we are able to get a Yes or No from the option, it should suffice. A couple of figures to further make it clear. Statement 1: The diameter of any circle that passes through points (2,4) and (5,10) will be equal to or more than root 45 i.e. 6. something. So radius of the circle will definitely be more than 3. We get a definite Yes from the option. Sufficient. Attachment:
Ques1.jpg [ 10.71 KiB  Viewed 4625 times ]
Statement 2: The diameter of any circle that passes through points (2,4) and (4,1) will be equal to or more than root 13 i.e. 3. something. So radius of the circle will definitely be more than 1.5. It could be less than 3, it could also be more than 3. We do not get a Yes or No. Either is possible. Hence not sufficient. Attachment:
Ques.jpg [ 10.22 KiB  Viewed 4625 times ]
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Re: A DS problem [#permalink]
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19 Nov 2010, 10:57
Hi, Thank you so much Bunuel & Karishma for the clearing this doubt.
I Feel relieved.



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Re: Is the radius of circle >3? [#permalink]
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14 Aug 2012, 07:38
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conty911 wrote: Is the radius of circle >3? Statement 1.) (2,4) & (5,10) are points on the circumference of circle. Statement 2.) (2,4) & (4,1) are points on the circumference of circle. Statement 1: I think we shoud use concepts of chords and calculate dist. of chords using coordinates. calculating distance from stmnt 1 = sqrt(45)=6.5 6.5/2=3.25 >3 (sufficient, since Diameter is the largest cord in the circle, any other chord must be smaller then it.Here the chord distance clearly exceeds the "diameter required =6" , which implies the given circle has diameter >=6.5 )
Stmnt2 distance =sqrt(13)=~3.6 , so here 3.6/2=1.8 <3
But here we are not sure about the answer since, there could be multiple chords with distance >3.6 Any side of a triangle is less than or equal to the diameter of the circumscribed circle. The diameter is the largest chord in any circle. Only for right angled triangles, one of the sides is equal to the diameter, specifically the hypotenuse. The question asks whether the diameter is greater than 6. (1) The distance between the two points is \(\sqrt{(52)^2+(104)^2}=\sqrt{9+36}>6.\) Sufficient. (2) The distance between the given two points is \(\sqrt{(41)^2+(24)^2}=\sqrt{9+4}<6.\) Since through two points there are infinitely many circles passing through, some of them will have diameter greater than 6, others won't. Not sufficient. Answer A
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Is the radius of the circle greater than 3? [#permalink]
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Re: Is the radius of the circle greater than 3? [#permalink]
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03 Mar 2015, 10:01
Great question. Thank you Bunuel.
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