It is currently 15 Dec 2017, 04:28

# Decision(s) Day!:

CHAT Rooms | Wharton R1 | Stanford R1 | Tuck R1 | Ross R1 | Haas R1 | UCLA R1

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the coordinate plane, a circle has center (2, -3) and

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 17 Aug 2009
Posts: 227

Kudos [?]: 301 [3], given: 25

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

### Show Tags

16 Dec 2009, 03:20
3
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

78% (01:04) correct 22% (01:04) wrong based on 1482 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Apr 2012, 00:35, edited 1 time in total.
Edited the question and added the OA

Kudos [?]: 301 [3], given: 25

Manager
Joined: 09 May 2009
Posts: 204

Kudos [?]: 271 [2], given: 13

Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

16 Dec 2009, 03:24
2
KUDOS
IMO E ---> 18 pi

r=sqrt[{5-2}^2+{0+3}^2]=3 sqrt2.......(dist formula)
area=pi r^2
=pi* (3 sqrt2)^2=18 pi
_________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Kudos [?]: 271 [2], given: 13

Senior Manager
Joined: 30 Aug 2009
Posts: 281

Kudos [?]: 194 [1], given: 5

Location: India
Concentration: General Management
Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

16 Dec 2009, 08:51
1
KUDOS
E - 18pi

r^2 = [(5-2)^2 + (-3-0)^2] = 18. So Area = pi * r^2 = 18pi

Kudos [?]: 194 [1], given: 5

Manager
Joined: 17 Aug 2009
Posts: 227

Kudos [?]: 301 [1], given: 25

Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

17 Dec 2009, 03:55
1
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)

Hope this helps

Consider Kudos for the post

Kudos [?]: 301 [1], given: 25

Manager
Joined: 24 Jul 2009
Posts: 192

Kudos [?]: 47 [1], given: 10

Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

09 Jan 2010, 22:52
1
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?
_________________

Reward wisdom with kudos.

Kudos [?]: 47 [1], given: 10

Manager
Joined: 24 Jul 2009
Posts: 192

Kudos [?]: 47 [0], given: 10

Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

09 Jan 2010, 23:00
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.
_________________

Reward wisdom with kudos.

Kudos [?]: 47 [0], given: 10

Manager
Joined: 10 Nov 2010
Posts: 244

Kudos [?]: 421 [0], given: 22

Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

02 Apr 2012, 00:01
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 proof of understanding is the ability to explain it.

Kudos [?]: 421 [0], given: 22

Math Expert
Joined: 02 Sep 2009
Posts: 42620

Kudos [?]: 135707 [16], given: 12706

Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

02 Apr 2012, 00:34
16
KUDOS
Expert's post
6
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 31420 times ]
The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: $$r^2=3^2+3^2=18$$ --> $$area=\pi{r^2}=18\pi$$.

Hope it helps.
_________________

Kudos [?]: 135707 [16], given: 12706

Manager
Joined: 28 Jul 2011
Posts: 233

Kudos [?]: 164 [1], given: 16

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

### Show Tags

04 Apr 2012, 04:26
1
KUDOS
E

we get height as "3" using points (2,0) & (2,-3)

we get base as "3" using points (2,0) & (5,0)

therefore (radius)^2 = (3)^2 + (3)^2

area = pi * r*r
=18pi

Kudos [?]: 164 [1], given: 16

Manager
Joined: 07 May 2013
Posts: 107

Kudos [?]: 30 [0], given: 1

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

### Show Tags

10 Nov 2013, 17:04
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.

Kudos [?]: 30 [0], given: 1

Manager
Joined: 12 Jan 2013
Posts: 217

Kudos [?]: 83 [0], given: 47

Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

19 Dec 2013, 00:13
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: $$r^2=3^2+3^2=18$$ --> $$area=\pi{r^2}=18\pi$$.

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?

Kudos [?]: 83 [0], given: 47

Math Expert
Joined: 02 Sep 2009
Posts: 42620

Kudos [?]: 135707 [0], given: 12706

Re: Coordinate Geometry from Paper test [#permalink]

### Show Tags

19 Dec 2013, 00:17
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: $$r^2=3^2+3^2=18$$ --> $$area=\pi{r^2}=18\pi$$.

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 $$area=\pi{r^2}=18\pi$$.
_________________

Kudos [?]: 135707 [0], given: 12706

Intern
Joined: 25 Jun 2014
Posts: 2

Kudos [?]: 1 [1], given: 7

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

### Show Tags

25 Jun 2014, 21:30
1
KUDOS
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.

Kudos [?]: 1 [1], given: 7

Intern
Joined: 08 Nov 2015
Posts: 35

Kudos [?]: 2 [0], given: 24

Schools: Pepperdine '19
Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

### Show Tags

01 Feb 2016, 12:07
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

Kudos [?]: 2 [0], given: 24

Current Student
Joined: 23 Mar 2016
Posts: 37

Kudos [?]: 6 [0], given: 0

Schools: Tulane '18 (M)
Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

### Show Tags

02 May 2016, 14:32
given: center = (2,-3)
passes through (5,0) <<< means that a point of the circle, therefore the edge of the circle

let's make a triangle from the center to (5,0)

side 1 = 3, side 2 = 3, radius = x
use pythag (we have created a right triangle at the x axis)
3^2+3^2=x^2
9+9=18
3^2+3^2=√18
A=πr^2
A=π(√18)^2
A=18π

Kudos [?]: 6 [0], given: 0

Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1810

Kudos [?]: 981 [1], given: 5

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

### Show Tags

03 May 2016, 04:44
1
KUDOS
Expert's post
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

Because the circle passes through point (5,0) and has center (2,-3), we know that the distance between these points is the radius of the circle.

Since we are given the coordinates for each point, the easiest thing to do is to use the distance formula to determine the circle's radius. The distance formula is:

Distance= √[(x2 – x1)^2 + (y2 – y1)^2]

We are given two ordered pairs, so we can label the following:

x1 = 2
x2 = 5

y1 = -3
y2 = 0

When we plug these values into the distance formula, we have:

Distance= √[(5 – 2)^2 + (0 – (-3))^2]

Distance= √ [(3)^2 + (3)^2]

Distance= √ [9 + 9]

Distance = √ [18]

Distance = √9 x √2

Distance = 3 x √2

Thus, we know that the radius = 3 x √2.

Finally, we can use the radius to determine the area of the circle.

area = ∏r^2

area = ∏(3 x √2 )^2

area = ∏(9 x 2 )

area = 18∏

_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 981 [1], given: 5

Senior Manager
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 446

Kudos [?]: 524 [0], given: 59

Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
Re: In the coordinate plane, a circle has center (2, -3) and [#permalink]

### Show Tags

16 May 2016, 19:44
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-16 at 8.40.06 PM.png [ 143.81 KiB | Viewed 2129 times ]

_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and online via Skype, since 2002.

GMAT Action Plan - McElroy Tutoring

Kudos [?]: 524 [0], given: 59

SVP
Joined: 06 Nov 2014
Posts: 1903

Kudos [?]: 550 [0], given: 23

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

### Show Tags

10 Jul 2016, 19:57
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

since the circle passes through (5,0) the radius will be the line joining the centre (2,-3) and (5,0)
Length of radius = $$\sqrt{{3^2+3^2}}$$ = $$\sqrt{{18}}$$

Area = π*r^2 = π*18

Correct Option: E

Kudos [?]: 550 [0], given: 23

SVP
Joined: 11 Sep 2015
Posts: 1910

Kudos [?]: 2752 [1], given: 364

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

### Show Tags

25 Nov 2017, 14:04
1
KUDOS
Expert's post
Top Contributor
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

IMPORTANT: The distance from the center to a point ON the circle = the radius (that's the definition of "radius")
What is the distance from (2, -3) to (5, 0)?

If we apply the distance formula, we get:
Distance = √[(2 - 5)² + (-3 - 0)²]
= √[(-3)² + (-3)²]
= √[9 + 9]
= √18

So, the radius is √18

Area of circle = π(radius
= π(√18
= 18π
= E

Cheers,
Brent
_________________

Brent Hanneson – Founder of gmatprepnow.com

Kudos [?]: 2752 [1], given: 364

VP
Joined: 22 May 2016
Posts: 1131

Kudos [?]: 404 [0], given: 646

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

### Show Tags

25 Nov 2017, 15:13
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

To each her own; I find the distance formula cumbersome, especially given a center and a point.

I used the general equation for a circle (also called the center-radius form), where $$(h,k)$$ are the center's coordinates:

$$(x - h)^2 + (y - k)^2 = r^2$$

$$h = 2, k = -3$$; this circle's general equation is

$$(x - 2)^2 + (y + 3)^2 = r^2$$

Take the point given, (5,0), and plug coordinates into the equation (5 for x, 0 for y):

$$(5 - 2)^2 + (0 + 3)^2 = r^2$$
$$3^2 + 3^2 = r^2$$
$$(9 + 9) = 18 = r^2$$

Stop there.* Circle's area = $$\pi r^2$$, and $$r^2 = 18$$

So area = $$\pi r^2 = 18\pi$$

* If you needed to find the radius:
$$18 = r^2$$
$$\sqrt{r^2}=\sqrt{9*2}$$
$$r = 3\sqrt{2}$$

Kudos [?]: 404 [0], given: 646

In the coordinate plane, a circle has center (2, -3) and   [#permalink] 25 Nov 2017, 15:13
Display posts from previous: Sort by

# In the coordinate plane, a circle has center (2, -3) and

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.