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.

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:

The distance between points is sqrt10, sqrt10, sqrt20. This is 45, 45, 90 triangle. By circle properties if any inscribed triangle has 90 degree angle its hypotenuse passes through its center. So diameter is hypotenuse of triangle =sqrt20.

My approach would be: Let x,y is the center of the circle, then distance between center and any point on the circle is same equal to R. Given that (1,2), (2,5) and (5,4) lie on circle, we can have following equations:

(x-1)^2 + (y-2)^2 = (x-5)^2 +(y-4)^2 solve it for 2x+y=9

Next take (2,5) and (5,4) (x-2)^2 + (y-5)^2 = (x-5)^2 +(y-4)^2 and you will get 3x-y=6

Solve for x and y center is (3,3) and distance i.e. Radius is \sqrt{(5-3)^2+(4-3)^2} = \sqrt{5}

So Diameter is \sqrt{20}

I did the same. However it's way too long and complex calculation, gmat suggests easier solutions to be within 2 minutes limit

Just want to note that I've seen a lot of people throwing around the "rule" that if a triangle is inscribed in a circle, then it either a) must be a right triangle or b) have it's hypotenuse pass through the circle's center. Neither of these are true...

The rule is that IF a right triangle is inscribed in a circle, then it's hypotenuse passes through the center of the circle (and is thus a diameter). You can inscribe a virtually infinite number of triangles in a circle which would not be right triangles and would thus not pass through the circle's center.

So to do this problem you must first determine that the triangle formed by the points is indeed a right triangle. Using (Y1-Y2)/(X1-X2) to find the slopes of the three line segments, you will see that segments intersecting at (2,5) have inverse and opposite slopes (3, -1/3) and are thus perpendicular to each other and meet at a right angle. From there it is as simple as applying the pythagorean theorm to find the long side, in this case x^2 = 4^2 + 2^2.

It seems like some people may have gotten "lucky" assuming that the long side was the hypotenuse of a right triangle, and thus a diameter (and the answer) - Just wanted to note that wouldn't always be the case
_________________

Good Question. For a sec I thought it is a tough question till I worked out the length of the sides.

The length of the sides work out to \sqrt{10} \sqrt{10} and \sqrt{20}

It implies that it is an isosceles triangle with a right angle. All triangles with base as diameter and the vertex touching a point on circumference of the circle - the vertex angle is 90 degree.

So it works out. The base with side \sqrt{20} is the diameter of the circle as well.

Guys, correct me if I am wrong in my logic. Ignore the point (2,5). Considering the fact that the circle touches all the points, the line joining the (1,2)--say point A and (5,4)---say point B is the diameter(AB). Now, draw perpendiculars from A and B across the coordinates which will join at Point C---(5,2). Triangle, thus, formed is right angled triangle- triangle ABC. Using simple coordinate numbers, the length of AC = 4 and length of BC = 2. Applying Pythogoras Theorem, the diameter i.e. the hypotenuse of the triangle ABC = sqrt (20).

draw the above figure with simple coordinates to follow the above logic.

this is exactly how i did it.

Its much easier to visual if you have either a horizontal and/or vertical line in your triangle.

I drew dotted lines from points (5,4) and (1,2) to point (5,2) creating a right triangle angle at that new point with a shared hypotenuse with the original triangle. The length of the shared hypotenuse is what we are looking for.

side x which goes from point (1,2) to (5,2) has length 4 side y which goes from point (5,4) to (5,2) has length 2

a^2 + B^2 = c^2 c is the value we are looking for as length c is equal to side z the shared hypotenuse.