Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 May 2013, 19:05

Circle passes through points (1, 2), (2, 5), and (5, 4).

Author Message
TAGS:
Manager
Joined: 12 Feb 2006
Posts: 119
Followers: 1

Kudos [?]: 0 [0], given: 0

Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink]  12 Feb 2007, 01:00
00:00

Question Stats:

82% (03:18) correct 17% (01:48) wrong based on 6 sessions
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}
[Reveal] Spoiler: OA
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8

Kudos [?]: 71 [1] , given: 0

1
KUDOS
(B) for me

The equation of circle C in the XY plan is :
(x-a)^2 + (y-b)^2 = r^2

Where :
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:
(a-1)^2 + (b-2)^2 = r^2
<=> a^2 -2*a + 1 + b^2 - 4*b + 4 = r^2 (1)

(2, 5) on C implies:
(a-2)^2 + (b-5)^2 = r^2
<=> a^2 -4*a + 4 + b^2 - 10*b + 25 = r^2 (2)

(5, 4) on C implies:
(a-5)^2 + (b-4)^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):
(a-1)^2 + (b-2)^2 = r^2
<=> (2)^2 + (1)^2 = r^2
<=> r = sqrt(5)

Thus,
D = 2*sqrt(5) = sqrt(20)
Senior Manager
Joined: 23 Jun 2006
Posts: 398
Followers: 1

Kudos [?]: 293 [0], given: 0

Re: Cord Geometry [#permalink]  12 Feb 2007, 09:53
kevincan wrote:
bz9 wrote:
These types of question really baffle me. I know Fig is pretty wicked with the graph paper so any help would be appreciated.

Circle passes through points A(1, 2), B(2, 5), and C(5, 4). What is the diameter of the circle?

a. sqrt(18)
b. sqrt(20)
c. sqrt(22)
d. sqrt(26)
e. sqrt(30)

Yipes! What do you notice about the slopes of line segments AB and BC?

yeah.... that's what i'd do. one should notice that the triangle ABC has right angle (in B), which means thata AC is the diameter. than it is easy to compute the distance between two points... (getting same result as Fig)
Manager
Joined: 12 Feb 2006
Posts: 119
Followers: 1

Kudos [?]: 0 [0], given: 0

Ok,

If I'm doing this right, the slope of A, B is 3 and B, C is -1/3 i.e. perpendicular bisector. Therefore, as hobbit pointed out, B is a right trangle. Then use P theorm to solve.

Thanks a lot guys. Makes total sense.
Intern
Joined: 10 Jun 2009
Posts: 33
Location: Stockholm, Sweden
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Cord Geometry [#permalink]  16 Jun 2009, 06:04
1. Since the circle has an inscribed square we first have to compute the lenght of the side of the square which is 1^2 + 3^2 = 10.
2. After that we have to calculate the diagonal of the square which is 10 (the squared root of 10) + 10 = root 20.
Current Student
Joined: 12 Jun 2009
Posts: 1847
Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)
Followers: 20

Kudos [?]: 176 [0], given: 52

Re: Cord Geometry [#permalink]  16 Jun 2009, 08:22
yeah just use the distance formula and you can see it is same sides(right triangle) so 1-1-sqrt(2) you get sqrt(2)*sqrt(10) = sqrt(20)
_________________

Manager
Joined: 22 Jul 2009
Posts: 200
Followers: 2

Kudos [?]: 141 [0], given: 18

Re: Cord Geometry [#permalink]  19 Aug 2009, 20:10
Diameter = distance between (1,2) and (5,4) = sqrt [ (5-1)^2 + (4-2)^2 ] = sqrt [ 16 + 4 ] = sqrt [20]

Simple is good.
Manager
Joined: 22 Sep 2009
Posts: 228
Location: Tokyo, Japan
Followers: 2

Kudos [?]: 14 [0], given: 8

Re: Cord Geometry [#permalink]  25 Nov 2009, 01:05
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?
Manager
Joined: 24 Jul 2009
Posts: 197
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Followers: 4

Kudos [?]: 17 [0], given: 10

Re: Cord Geometry [#permalink]  10 Jan 2010, 01:05
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.
_________________

Reward wisdom with kudos.

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11611
Followers: 1801

Kudos [?]: 9597 [2] , given: 829

Re: Cord Geometry [#permalink]  10 Jan 2010, 09:59
2
KUDOS
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 [ 6.47 KiB | Viewed 3744 times ]

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 XY-plane. For more please see the Triangles and Circles chapters of Math Book in my signature.

Hope it's clear.
_________________
Manager
Joined: 24 Jul 2009
Posts: 197
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Followers: 4

Kudos [?]: 17 [0], given: 10

Re: Cord Geometry [#permalink]  11 Jan 2010, 21:26
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 XY-plane. For more please see the Triangles and Circles chapters of Math Book in my signature.

Hope it's clear.

Thanks for the explanation. Reworked the problem. Got the negative reciprocal slopes. Did distance formula. Now see how I can get the correct answer in 3 quick steps. Appreciate the lesson learned about perpendicular line segments.
_________________

Reward wisdom with kudos.

Intern
Joined: 07 Aug 2012
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Cord Geometry [#permalink]  10 Aug 2012, 07:03
I dont know how you could assuse AC is the diameter. Because there is nothing to indicate that AC goes through the center right? I am a little confused about the solution.
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11611
Followers: 1801

Kudos [?]: 9597 [0], given: 829

Re: Cord Geometry [#permalink]  10 Aug 2012, 07:43
pramoddikshith wrote:
I dont know how you could assuse AC is the diameter. Because there is nothing to indicate that AC goes through the center right? I am a little confused about the solution.

AC is the diameter because angle at B is 90 degrees. Check this post: circle-passes-through-points-1-2-2-5-and-42105.html#p672194
_________________
Senior Manager
Joined: 15 May 2011
Posts: 277
Location: Costa Rica
GMAT 1: 640 Q V0
GMAT 2: 710 Q42 V45
GMAT 3: 740 Q48 V42
GMAT 4: Q V0
GMAT 5: Q V0
GPA: 3.3
WE: Research (Consulting)
Followers: 7

Kudos [?]: 47 [0], given: 134

Re: Cord Geometry [#permalink]  11 Aug 2012, 15:33
Bunuel wrote:
pramoddikshith wrote:
I dont know how you could assuse AC is the diameter. Because there is nothing to indicate that AC goes through the center right? I am a little confused about the solution.

AC is the diameter because angle at B is 90 degrees. Check this post: circle-passes-through-points-1-2-2-5-and-42105.html#p672194

Bunuel is this particular property tested often on the GMAT, not the right triangle one, but the formular mentioned at the top:
The equation of circle C in the XY plan is :
(x-a)^2 + (y-b)^2 = r^2

_________________

Work experience (as of June 2012)
2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst)
2 yrs - Advertising industry (client servicing)

SVP
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1756
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 50

Kudos [?]: 145 [0], given: 108

Re: Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink]  25 Aug 2012, 20:05
+1 B

The angle in the point (2; 5) is right. You can calculate that evaluating the slopes of the lines formed by the points.
With that angle, we can know that the line formed by the points (1;2) and (5;4) is the diameter of the circle.
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: my-ir-logbook-diary-133264.html

Find out what's new at GMAT Club - latest features and updates

Re: Circle passes through points (1, 2), (2, 5), and (5, 4).   [#permalink] 25 Aug 2012, 20:05
Similar topics Replies Last post
Similar
Topics:
Line A and B pass through the point (16/5,12/5). What is the 1 06 May 2006, 00:27
Line L passes through points A (5,0) and B (0,2), O is the 1 09 Aug 2006, 03:39
Circle passes through points (1,2) (2,5) (5,4) What is 4 17 Oct 2007, 12:18
Circle passes through points (1, 2) , (2, 5) 7 26 Aug 2008, 09:47
5 Line m and n pass through point (1,2). Is the slope of m 12 20 Feb 2010, 23:38
Display posts from previous: Sort by