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If a circle passes through points (1, 2), (2, 5), and (5, 4)

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If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink] New post 12 Feb 2007, 00:00
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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}
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Oct 2013, 02:51, edited 2 times in total.
Renamed the topic and edited the question.
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 [#permalink] New post 12 Feb 2007, 01:18
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(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)
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Re: Cord Geometry [#permalink] New post 12 Feb 2007, 08:53
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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)
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 [#permalink] New post 12 Feb 2007, 20:41
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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.
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Re: Cord Geometry [#permalink] New post 16 Jun 2009, 05:04
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.
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Re: Cord Geometry [#permalink] New post 16 Jun 2009, 07:22
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)
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Re: Cord Geometry [#permalink] New post 19 Aug 2009, 19:10
Diameter = distance between (1,2) and (5,4) = sqrt [ (5-1)^2 + (4-2)^2 ] = sqrt [ 16 + 4 ] = sqrt [20]

Simple is good.
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Re: Cord Geometry [#permalink] New post 25 Nov 2009, 00:05
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?
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Re: Cord Geometry [#permalink] New post 10 Jan 2010, 00:05
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.
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Re: Cord Geometry [#permalink] New post 10 Jan 2010, 08:59
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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
Math_Tri_inscribed.png [ 6.47 KiB | Viewed 6490 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.
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Re: Cord Geometry [#permalink] New post 11 Jan 2010, 20:26
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.
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Re: Cord Geometry [#permalink] New post 10 Aug 2012, 06:03
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.
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Re: Cord Geometry [#permalink] New post 10 Aug 2012, 06:43
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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
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Re: Cord Geometry [#permalink] New post 11 Aug 2012, 14:33
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

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Re: Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink] New post 25 Aug 2012, 19:05
+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.
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Re: Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink] New post 13 Oct 2013, 02:14
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink] New post 13 Oct 2013, 02:51
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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}

Look at the diagram below:
Image
Calculate the lengths of the sides of triangle ABC:
AB=\sqrt{10};
BC=\sqrt{10};
AC=\sqrt{20}=\sqrt{2}*\sqrt{10};

As we see the ratio of the sides of triangle ABC is 1:1:\sqrt{2}, so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2}).

So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20}.

Answer: B.
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink] New post 11 Feb 2014, 02:55
Bunuel 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}

Look at the diagram below:
Image
Calculate the lengths of the sides of triangle ABC:
AB=\sqrt{10};
BC=\sqrt{10};
AC=\sqrt{20}=\sqrt{2}*\sqrt{10};

As we see the ratio of the sides of triangle ABC is 1:1:\sqrt{2}, so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2}).

So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20}.

Answer: B.



Bunuel.. How did we get that ab and bc are square root 10?
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink] New post 11 Feb 2014, 03:05
Expert's post
sanjoo wrote:
Bunuel 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}

Look at the diagram below:
Image
Calculate the lengths of the sides of triangle ABC:
AB=\sqrt{10};
BC=\sqrt{10};
AC=\sqrt{20}=\sqrt{2}*\sqrt{10};

As we see the ratio of the sides of triangle ABC is 1:1:\sqrt{2}, so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2}).

So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20}.

Answer: B.



Bunuel.. How did we get that ab and bc are square root 10?


Check Distance between two points here: math-coordinate-geometry-87652.html
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink] New post 11 Feb 2014, 05:54
Ohkey Bunuel.. Thanks alot ! now question becomes easy to solve ! It has been long time i haven't touch coordinate geometry !
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)   [#permalink] 11 Feb 2014, 05:54
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