Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 13038 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 _________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

Re: Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink]

Show Tags

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: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

Re: Circle passes through points (1, 2), (2, 5), and (5, 4). [#permalink]

Show Tags

13 Oct 2013, 03:14

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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]

Show Tags

11 Feb 2014, 03:55

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 !

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?

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...