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:

yeah nice approach, but how did u spot the 90 degrees?

I think he spotted it by drawing a diagram without the circle. Or not.

Let's start with (1,2) and (2,5). Draw a right triangle keeping the X-coordinate constant, and we have a third point at (1,5). This gives us three legs of a right triangle. Side A=(1,2) to (1,5) = 3, side B=(1,5) to (2,5) = 1, side C=(2,5) to 1,2)=\(sqrt(A^2+B^2)=sqrt(1^2+3^2)=sqrt(10)\). Use the same method for the points (2,5) and (5,4), but holding the Y-coordinate constant, and the hypotenuse is also \(sqrt(10)\). Applying the same method for points (5,4) and (1,2) you find that the hypotenuse for that triangle is \(sqrt(20)\). Looking at the triangle with the lengths of the legs and hypotenuse calculated you can see that the triangle is a right triangle.

yeah nice approach, but how did u spot the 90 degrees?

if you see the two sides are sqrt(10),sqrt(10) and third side is sqrt (20) they are in the ratio 1:1:sqrt(2) .. this possible only for right angle traingle.

this large side must be diameter.

Great job!! Mohindru.
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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

What is the diameter of the circle? * \(\sqrt{18}\) * \(\sqrt{20}\) * \(\sqrt{22}\) * \(\sqrt{26}\) * \(\sqrt{30}\)

i see no easy way of solving this based on teh OA. any thoughts?

The three points form a right angle triangle at point (2,5) .

Distance between (1,2) and (2,5) = \(sqrt (10)\) (1,2) and (5,4) = \(sqrt (20)\) and (2,5) and (5,4) = \(sqrt(10)\)

Therefore diameter of circle = hypoteneuse of triangle = \(sqrt(20)\)

Since any right angle triangle formed between three points of circle hypoteneuse is diameter

Its nice that you were able to spot the hypoteneuse and the point of right angle. However with this approach, you would be stuck if this is not the case. The following is an approach which would give result no matter where the points are on the circle. The drawback is that it would take 3-5 min to solve.

Equation of circle is

(x-h)^2 + (y-k)^2 = r^2 where (h,k) are coordinates of center, r = radius and (x,y) are points on circle

Insert the 3 points to obtain the following equations