Coordinate Geometry.......

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pzazz12: Master | Next Rank: 500 Posts; Posts: 155; Joined: Wed Sep 22, 2010 3:54 am; Thanked: 6 times

by pzazz12 » Fri Oct 01, 2010 1:15 am

What is the equation of a circle of radius 6 units centered at (3, 2)?

A. x2 + y2 + 6x - 4y = 23
B. x2 + y2 - 6x + 4y = 23
C. x2 + y2 + 6x + 4y = 23
D. x2 + y2 - 6x - 4y = - 23
E. x2 + y2 - 6x - 4y = 23

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Source: Beat The GMAT — Problem Solving | Fri Oct 01, 2010 1:15 am

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Fri Oct 01, 2010 1:30 am

pzazz12 wrote:What is the equation of a circle of radius 6 units centered at (3, 2)?

A. x2 + y2 + 6x - 4y = 23
B. x2 + y2 - 6x + 4y = 23
C. x2 + y2 + 6x + 4y = 23
D. x2 + y2 - 6x - 4y = - 23
E. x2 + y2 - 6x - 4y = 23

eqtn of a circle with radius r and center at (a,b) is
(x-a)^2 + (y-b)^2 = r^2

Hence, (x-3)^2 + (y-2)^2 = 6^2
x^2 -6x + 9 + y^2 -4y +4 = 36
x^2 + y^2 -6x -4y + 13 = 36
x^2 + y^2 -6x -4y =23

pick E.

Hope this helps!!

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don't be afraid of failure and don't abandon it.
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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Oct 01, 2010 2:46 am

pzazz12 wrote:What is the equation of a circle of radius 6 units centered at (3, 2)?

A. x2 + y2 + 6x - 4y = 23
B. x2 + y2 - 6x + 4y = 23
C. x2 + y2 + 6x + 4y = 23
D. x2 + y2 - 6x - 4y = - 23
E. x2 + y2 - 6x - 4y = 23

Point (3,-4) is on the circle because it is 6 units from (3,2). Plug x=3, y=-4 into the answer choices.

Only E works:
(3)^2 + (-4)^2 - (6*3) - (4*-4) = 23
9 + 16 - 18 - (-16) = 23
23=23.

Please note that this sort of question will not appear on the GMAT. The equation of a circle is beyond the scope of the test.

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