May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 06 Aug 2011
Posts: 329

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
11 Feb 2014, 08:11
Thanks again Bunuel
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Intern
Joined: 05 Feb 2014
Posts: 42

Re: Cord Geometry
[#permalink]
Show Tags
17 Jun 2014, 06:57
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 ?



Math Expert
Joined: 02 Sep 2009
Posts: 55267

Re: Cord Geometry
[#permalink]
Show Tags
17 Jun 2014, 07:45
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#p672194
_________________



Intern
Joined: 05 Feb 2014
Posts: 42

Re: Cord Geometry
[#permalink]
Show Tags
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 ?



Math Expert
Joined: 02 Sep 2009
Posts: 55267

Re: Cord Geometry
[#permalink]
Show Tags
18 Jun 2014, 08:51
gauravsoni wrote: Bunuel wrote: gauravsoni wrote: 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 ? I guess you did not read that post to the end: 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.
_________________



Intern
Joined: 16 May 2015
Posts: 2
Location: India
Concentration: Operations, Finance
GPA: 4
WE: Other (Other)

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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.



Manager
Joined: 28 Apr 2016
Posts: 87

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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)



Manager
Joined: 05 Sep 2014
Posts: 73

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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



Senior Manager
Status: You have to have the darkness for the dawn to come
Joined: 09 Nov 2012
Posts: 288
Daboo: Sonu
GMAT 1: 590 Q49 V20 GMAT 2: 730 Q50 V38

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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
_________________
You have to have the darkness for the dawn to come.
Give Kudos if you like my post



Intern
Joined: 24 Oct 2014
Posts: 15
Location: Spain

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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.



Director
Joined: 31 Jul 2017
Posts: 516
Location: Malaysia
GPA: 3.95
WE: Consulting (Energy and Utilities)

If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!



Intern
Joined: 01 Feb 2018
Posts: 18

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
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?



Intern
Joined: 02 Feb 2018
Posts: 36

Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
Show Tags
11 Nov 2018, 09:35
Fernandocma wrote: 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. Agree, that's also what I thought




Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)
[#permalink]
11 Nov 2018, 09:35



Go to page
Previous
1 2
[ 33 posts ]



