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:

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

17 Dec 2009, 04:55

1

This post received KUDOS

OA is E

The distance formula in coordinate geometry is used to calculate the distance between 2 points whose coordinates are given

Lets say we have to calculate the distance between 2 points (x1, y1) and (x2, y2)

It is given by -

sqrt [ (x2-x1)^2 + (y2-y1)^2]

In this case since we are given the coordinates of the center and the fact that the circle passes thru (5,0), we can calculate the radius (needed for finding the area of the circle), by calculating the distance from the center to point (5,0)

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

09 Jan 2010, 23:52

1

This post received KUDOS

Got E. Used distance formula. Didn't solve for hypotenuse since it was an isoceles right triangle. Plugged in 3 sqrt 2 for radius in area formula. Thanks for the post. Got the answer pretty quickly. What level would you say this is? _________________

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

10 Jan 2010, 00:00

gottabwise wrote:

Got E. Used distance formula. Didn't solve for hypotenuse since it was an isoceles right triangle. Plugged in 3 sqrt 2 for radius in area formula. Thanks for the post. Got the answer pretty quickly. What level would you say this is?

Just copied problem into notes and recognized how I did extra work...should've just did distance formula b/w (5,0) and (2,-3)...r^2=(5-2)^3+(0--3)^2=sqrt18=3sqrt2. I guess that's why I'm reviewing right now. Noticing that r^2 doesn't need to be solved for either. _________________

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

02 Apr 2012, 01:34

9

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

GMATD11 wrote:

zaarathelab wrote:

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?

A. 3π B. 3√2π C. 3√3π D. 9π E. 18π

Can any body draw the picture for this

i thought y coordinate between (5,0) & (2,-3) will be radius i.e 3 and area will be 3*3pie

The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below:

Attachment:

graph.png [ 10.12 KiB | Viewed 17544 times ]

The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) --> \(area=\pi{r^2}=18\pi\).

Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

Show Tags

10 Nov 2013, 18:04

The equation of a circle is given by the formula \((x-a)^2+(y-b)^2=r^2\) where a and b represent the coordinates of the centre and r the radius. Substituting we get \((x-2)^2+(y+3)^2=r^2\). Now, given that the circle passes through the points (5,0). Hence, the when we substitute x=5 and y=0 in the equation above it should be satisfied. Substituting we get \((5-2)^2+(0+3)^2=r^2\). Therefore \(r^2=18\). Area=pi*r^2=18pi.

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

19 Dec 2013, 01:13

Bunuel wrote:

GMATD11 wrote:

zaarathelab wrote:

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?

A. 3π B. 3√2π C. 3√3π D. 9π E. 18π

Can any body draw the picture for this

i thought y coordinate between (5,0) & (2,-3) will be radius i.e 3 and area will be 3*3pie

The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below:

Attachment:

graph.png

The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) --> \(area=\pi{r^2}=18\pi\).

Answer: E.

Hope it helps.

But if the radius = hypothenuse, and if hypothenuse = 18, then the area is 18*18*pi, not 18*pi.. Right? Or am I missing something?

Re: Coordinate Geometry from Paper test [#permalink]

Show Tags

19 Dec 2013, 01:17

Expert's post

aeglorre wrote:

Bunuel wrote:

GMATD11 wrote:

Can any body draw the picture for this

i thought y coordinate between (5,0) & (2,-3) will be radius i.e 3 and area will be 3*3pie

The point is that the radius does not equal to 3, it equals to \(3\sqrt{2}\). You can find the length of the radius either with the distance formula (the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)) or with Pythagoras theorem. Look at the diagram below:

Attachment:

graph.png

The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: \(r^2=3^2+3^2=18\) --> \(area=\pi{r^2}=18\pi\).

Answer: E.

Hope it helps.

But if the radius = hypothenuse, and if hypothenuse = 18, then the area is 18*18*pi, not 18*pi.. Right? Or am I missing something?

No. Notice that we get that \(r^2=18\), not r. Thus \(area=\pi{r^2}=18\pi\). _________________

Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

Show Tags

25 Jun 2014, 22:30

I used Ballpark method : (x,y) = center (2,-3) to circumference of circle (5,0) that means radius should be more than 3. let's say 3+ area = π*r^2, should be more than π*3^2 ; answer > 9π only 18π is eligible.

Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

Show Tags

22 Jul 2015, 06:16

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. _________________

Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

Show Tags

01 Feb 2016, 13:07

using the general formula of a circle is the fastest method to solve this problem. (x-a)^2 + (y-b)^2 = r^2 where (a,b) is the coordinate from which the circle passes through

gmatclubot

Re: In the coordinate plane, a circle has center (2, -3) and
[#permalink]
01 Feb 2016, 13:07

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...