To find the points outside the circle

This topic has expert replies
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pm
Thanked: 11 times
Followed by:8 members

To find the points outside the circle

by gmattesttaker2 » Sun Jan 19, 2014 6:09 pm
Hello,

Can you please assist with this?

Which of the following points lies outside
circle x^2 + y^2 = 25?

(A) (4, 3)
(B) (-4, -3)
(C) (9, 6)
(D) (2, 2)
(E) (2, 4)

OA: C

Thanks a lot,
Sri

User avatar
Master | Next Rank: 500 Posts
Posts: 283
Joined: Sun Jun 23, 2013 11:56 pm
Location: Bangalore, India
Thanked: 97 times
Followed by:26 members
GMAT Score:750

by ganeshrkamath » Sun Jan 19, 2014 9:00 pm
gmattesttaker2 wrote:Hello,

Can you please assist with this?

Which of the following points lies outside
circle x^2 + y^2 = 25?

(A) (4, 3)
(B) (-4, -3)
(C) (9, 6)
(D) (2, 2)
(E) (2, 4)

OA: C

Thanks a lot,
Sri
The general equation for a circle with center (A,B) and radius r is:
(x-A)^2 + (y-B)^2 = r^2

Comparing this with the given equation, we get:
(A,B) = (0,0) and r = 5

The maximum value of x or y dimension is +/- 5.

The point (9,6) is clearly outside this limit.

Choose C

Cheers
Every job is a self-portrait of the person who did it. Autograph your work with excellence.

Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494

User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pm
Thanked: 448 times
Followed by:34 members
GMAT Score:650

by theCodeToGMAT » Sun Jan 19, 2014 10:23 pm
x^2 + y^2 = 25
the equation can be re-written as:
(x-0)^2 + (y-0)^2 = (5)^2
Here,
radius = 5
origin = (0,0)
Maximum magnitude of x & y points = 5

So, [spoiler]{C}[/spoiler]
R A H U L

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Jan 20, 2014 4:41 am
gmattesttaker2 wrote:Hello,

Can you please assist with this?

Which of the following points lies outside
circle x^2 + y^2 = 25?

(A) (4, 3)
(B) (-4, -3)
(C) (9, 6)
(D) (2, 2)
(E) (2, 4)

OA: C

Thanks a lot,
Sri
x² + y² = r² is the equation of a circle that is centered at the origin and has a radius of r.
Thus:
x² + y² = 25 is the equation of a circle that is centered at the origin and has a radius of 5.
Since r=5, any point that is more than 5 places from the origin will be OUTSIDE the circle.
Since answer choice C -- (9.6) -- is more than 5 places from the origin, this point lies outside the circle.

The correct answer choice is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Mon Jan 20, 2014 3:42 pm
Hi Sri,

Each of the other explanations in this thread has focused on the "math" (solving that the radius of the circle = 5). Here's another way to look at the "math" though:

Since we're asked which of the points is "outside" the circle, and we're given an equation to work with, then the points "inside" the circle can be "plugged in" to the equation and will be <= 25.

ANY co-ordinate that, when plugged in, is > 25 would be outside the circle.

A: 4^2 + 3^2 = 25
B: (-4)^2 + (-3)^2 = 25
C: 9^2 + 6^2 = 117
D: 2^2 + 2^2 = 8
E: 2^2 + 4^2 = 20

The only answer that is > 25 is C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image