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:

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

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 \(AC=diameter=\sqrt{20}\).

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.

we can solve this problem in many ways first by finding the equation of circle whose center's co-ordinate is (a,b) and radius is r (x-a)^2+(y-b)^2=r^2 here withe the given co-ordinates, we can find 3 equation and we have three variable. upon solving we get the co-ordinate of center= (3,3) Now we can use distance formula to solve the radius sqrt(a1-a2)^2 +(b1-b2)^2 = distance between tow numbers, where (a1, a2) and (b1, b2) are co-ordinates. Method:2 Any line which passes through the mid point and center of a circle is perpendicular . Here with the given co-ordinates, we can find the equations of chords. So we find the equation of perpendicular bisector of chord. by solving two perpendicular bisector's equation we can find the center of co-ordinate. then by using distance formula we can find the distance between center and any given point. this found distance is the radius 2* radius = diameter _________________

You have to have the darkness for the dawn to come.