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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)
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.
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.
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) _________________
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.
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. _________________
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 10161 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.
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. _________________
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.
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.
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.
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]
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]
13 Oct 2013, 02:14
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink]
13 Oct 2013, 02:51
4
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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:
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 \(AC=diameter=\sqrt{20}\).
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink]
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:
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? _________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4) [#permalink]
11 Feb 2014, 03:05
Expert's post
1
This post was BOOKMARKED
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:
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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