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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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11 Feb 2014, 08:11
Thanks again Bunuel
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Re: Cord Geometry
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17 Jun 2014, 07:45



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Re: Cord Geometry
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17 Jun 2014, 09:27
Bunuel wrote: gauravsoni wrote: lonewolf wrote: Good method. Simple and quick.
Just for refreshing memories... If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle.
Is the above statement always true? but how do we get to know that the right angle is formed ? Check here: ifacirclepassesthroughpoints1225and42105.html#p672194I understand that if a right triangle is formed then the side will be the diameter. But in the question how do we come to know that a right angle triangle is being formed ?



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Re: Cord Geometry
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18 Jun 2014, 08:51



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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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23 Sep 2015, 08:43
Hi Friends,
This Q can b solved under a minute, as the question asks for the diameter, we know that the value must be 2*radius i.e. and among the options mentioned only sqrt(20) can be identified as 2*sqrt(5). Hence option B
Hopes it makes sense.



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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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11 May 2016, 07:39
Since the formula is (xa)^2 + (yb)^2 = r^2. How did you take (a1)^2 + (b2)^2 = r^2? Fig wrote: (B) for me The equation of circle C in the XY plan is : (xa)^2 + (yb)^2 = r^2Where : o (a,b) is the center of the circle o r is the radius of the circle Knowing that, u can plug the coordonates of each points to get a, b and r. (1, 2) on C implies:(a1)^2 + (b2)^2 = r^2 <=> a^2 2*a + 1 + b^2  4*b + 4 = r^2 (1)(2, 5) on C implies:(a2)^2 + (b5)^2 = r^2 <=> a^2 4*a + 4 + b^2  10*b + 25 = r^2 (2)(5, 4) on C implies:(a5)^2 + (b4)^2 = r^2 <=> a^2 10*a + 25 + b^2 8*b + 16 = r^2 (3)(1)  (2)<=> 2*a  3 + 6*b 21 =0 <=> 2*a + 6*b = 24 <=> a + 3*b = 12 (4)(1)  (3)<=> 8*a 24 + 4*b  12 = 0 <=> 2*a + b = 9 (5)(5) 2*(4)<=> 5*b = 15 <=> b = 3 From (1), we find a : a = 12  3*b = 12  3*3 = 3 Still from (1): (a1)^2 + (b2)^2 = r^2 <=> (2)^2 + (1)^2 = r^2 <=> r = sqrt(5) Thus, D = 2*sqrt(5) = sqrt(20)



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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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23 Jul 2016, 08:49
Bunuel wrote: gottabwise wrote: lonewolf: I'd like to know the answer as well. Great question.
I got B. Draw a coord plane, plotted pts, draw right triangle. Used distance formula for each side. Thought about the radius formula for circumscribed circles. Nixed idea. Noticed isoceles right triangle and realized sqrt 20 was diameter b/c it's the hypotenuse. Ended there. Glad I tried the problem. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle. Attachment: Math_Tri_inscribed.png So: If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle. TRUE. As for the question: the slope of line segment: A(1,2) and B(2,5) is 3 AND the slope of line segment: B(2,5) and C(5,4) is 1/3, the slopes are negative reciprocals hence these line segments are perpendicular to each other. We have right triangle ABC, AC=hypotenuse=Diameter. Also one tip: any three points, which are not collinear, define the unique circle on XYplane. For more please see the Triangles and Circles chapters of Math Book in my signature. Hope it's clear. Hi Bunnuel, Thanks a lot for your explanation, really helped me. However just wanted to understand how do we calculate the diameter if its not the right angle triangle in a circle. Apologies if this sounds too basic. I made a graph of the same, unable to analyze what could be diameter in such question. Thanks in advance for your kind help. Regards Megha



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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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19 Mar 2017, 21:26
bz9 wrote: If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle?
A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) Let us consider the origin is at (x,y) so from distance formula (x1)^2+ (y2)^2 =r^2 ..................................1 (x2)^2+(y5)^2= r^2 ..................................2 (x5)^2+(x4)^2 =r^2 ..................................3 on solving 1 and 2 we get x+3y =12 ..................4 on solving 1 and 3 we get 2x+ 6y=9 ..................5 now solve 4 and 5 (x,y)= (3,3) and is the center Sqrt((31)^2+(32)^2)= sqrt( 4+ 1)= sqrt(5)= radius Diamiter = 2*radius = 2* sqrt(5) = sqrt(20) hence B
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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15 Jan 2018, 03:32
I have a question:
It is a PS question, so the stem has to be enough to answer the question. If ABC wasn't a right triangle, it wouldn't be possible.
Therefore, I believe checking whether there is or not a right angle is not necessary and we should jump straight to calculating the distance between (1,2) and (5,4) because it has to be the hypotenuse (ergo the diameter).
Do you agree?
Thank you.



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If a circle passes through points (1, 2), (2, 5), and (5, 4)
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25 Jan 2018, 13:18
bz9 wrote: If a circle passes through points \((1, 2)\), \((2, 5)\), and \((5, 4)\), what is the diameter of the circle?
A. \(\sqrt{18}\) B. \(\sqrt{20}\) C. \(\sqrt{22}\) D. \(\sqrt{26}\) E. \(\sqrt{30}\) \(1st Method:\)The equation of circle with \((h,k)\) as centre is given by, \((xh)^2 + (yk)^2 = r^2\) As per the given equation, \((1h)^2+(2k)^2 = r^2 1\) \((2h)^2 + (5k)^2 = r^22\) \((5h)^3 + (4k)^2 = r^23\) Solving above 3 equations we get, \((h,k) = (3,3)\). Therefore, \(D = \sqrt{20}.\) \(2nd Method:\)The Points A, B & C form a right angle traingle. Hnece, the diameter will be equal to the distance between the Points forming the Hypotenous.
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
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28 Mar 2018, 11:17
How do we exactly know that the line combining these two points go through the center? What if (in another question) it does not go through the center?




Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) &nbs
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28 Mar 2018, 11:17



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