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by fskilnik@GMATH » Thu Dec 16, 2010 12:58 pm
Hi there,

Any point (x,y) that belongs to the circle given satisfies the equation (x-0)^2 + (y-a)^2 = r^2 where r>0 is the radius, therefore:

b^2 + (c-a)^2 = r^2 = d^2 + (e-a)^2 and simplifying we have:

b^2 + c^2 - 2ac = d^2 +e^2 - 2ea.

That means LS (left side) - RS (right side) = 2a (c-e) .

Therefore SUPPOSING we can "trust" the figure given, we have

(1) a > 0
(2) c < e

and with those in mind, LS - RS < 0 and we are done. (Answer in this scenario: B)

Regards,
Fabio.

P.S.: of course the OA could be [spoiler](D)[/spoiler], if we agree that (d,e) and (b,c) could "change places" and/or a is not necessarily positive... you got the idea... (I would consider this, therefore I would click [spoiler](D)[/spoiler], by the way.)
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by N:Dure » Thu Dec 16, 2010 1:58 pm
fskilnik wrote:Hi there,

Any point (x,y) that belongs to the circle given satisfies the equation (x-0)^2 + (y-a)^2 = r^2 where r>0 is the radius, therefore:

b^2 + (c-a)^2 = r^2 = d^2 + (e-a)^2 and simplifying we have:

b^2 + c^2 - 2ac = d^2 +e^2 - 2ea.

That means LS (left side) - RS (right side) = 2a (c-e) .

Therefore SUPPOSING we can "trust" the figure given, we have

(1) a > 0
(2) c < e

and with those in mind, LS - RS < 0 and we are done. (Answer in this scenario: B)

Regards,
Fabio.

P.S.: of course the OA could be [spoiler](D)[/spoiler], if we agree that (d,e) and (b,c) could "change places" and/or a is not necessarily positive... you got the idea... (I would consider this, therefore I would click [spoiler](D)[/spoiler], by the way.)
Thanks Fabio for your reply! I get it right up till "b^2 + c^2 - 2ac = d^2 +e^2 - 2ea.". What I don't get is the LS-RS part = 2a(c-e), how did you get to that?
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by GMATGuruNY » Thu Dec 16, 2010 3:48 pm
N:Dure wrote:Image
To make the situation easier to see, plug in a=1.

Radius = distance from (b,c) to (0,1) = √(b-0)² + (c-1)²
Radius = distance from (d,e) to (0,1) = √(d-0)² + (e-1)²

Since all radii are equal:
√(b-0)² + (c-1)² = √(d-0)² + (e-1)²
(b-0)² + (c-1)² = (d-0)² + (e-1)²
b² + c² -2c + 1 = d² + e² - 2e + 1
b² + c² -2c = d² + e² - 2e

Looking at the equation above, if b² + c² = d² + e², then -2c = -2e and c=e.
But we can tell from the picture that e>c, which means that the value being subtracted from the right side of the equation (2e) is larger than the value being subtracted from the left side of the equation (2c). Thus, in order for the right side to equal the left side, d² + e² > b² + c².

The correct answer is B.
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by N:Dure » Thu Dec 16, 2010 5:58 pm
As always, thanks GMATGuru for your help!

to make sure, since e>c, d^2 + e^2 has to be more than b^2+c^2 in order to equalize the 2 equations (the ones= r^2)
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by GMATGuruNY » Thu Dec 16, 2010 6:02 pm
N:Dure wrote:As always, thanks GMATGuru for your help!

to make sure, since e>c, d^2 + e^2 has to be more than b^2+c^2 in order to equalize the 2 equations (the ones= r^2)
Correct!
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].
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by fskilnik@GMATH » Fri Dec 17, 2010 5:02 am
N:Dure wrote: Thanks Fabio for your reply! I get it right up till "b^2 + c^2 - 2ac = d^2 +e^2 - 2ea.". What I don't get is the LS-RS part = 2a(c-e), how did you get to that?
Hi N:Dure,

Sorry for the delay.

The answer is: (b^2 + c^2) - (d^2 +e^2) = 2ac - 2ea = LS - RS = 2a(c-e) ...

I hope things are clear, now.

Regards,
Fabio.

P.S.: Mitch´s solution is a little less general than mine, but exactly the same reasoning. Be sure you realize that, because this will mean you really understood both!
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