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# The graph of y = | x - 2 | -1 divides the area formed by the

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The graph of y = | x - 2 | -1 divides the area formed by the [#permalink]  05 Apr 2006, 12:24
The graph of
y = | x - 2 | -1
divides the area formed by the equation
y^2 + x^2 + 2y - 4x = 4
in two parts.

What's the ratio of the smaller part to the bigger part?
VP
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I will go for 1:2

Its a SWAG but heres my attempt

y^2 + x^2 + 2y - 4x = 4 represents a circle

(x-2)^2 + (y+1)^2 = 9 (added 5 on each side)

So the centre of teh circle is (2,-1) and the radius is 3

Circle general eq is (x-a)^2 +(y-b)^2 = r^2

where (a,b) is the centre and r is the radius

Now take the other eq y = | x - 2 | -1

=> (y+1)^2 = (x-2)^2
Squaring both sides
y^2 + 2y + 1 = x^2 +4x +4

Taeke this and substituet in the 1st circle......
VP
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I think it should be 1:3
y = |x-2|-1 slices away a Quarter of the circle (x-2)^2+(y+1)^2 = 9.. But this Q is
I don't think we will see such Qs on GMAT..
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VP
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this question is not for me. so passsssssssssssssssss.
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giddi77 wrote:
I don't think we will see such Qs on GMAT..

You probably won't, but you have to admit that if you presuppose
that the two graphs won't look too difficult (as in the real GMAT),
then it's by far not as hard as it at first might look.

Even more, if you achieve to imagine what the two graphs look
like, it's quite trivial. In this case I intentionally chose those
"easy" relationship between the graphs, because the question
was designed to confuse most of you, but of course not all

By the way, 1:3 is correct.
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ccax wrote:
giddi77 wrote:
I don't think we will see such Qs on GMAT..

You probably won't, but you have to admit that if you presuppose
that the two graphs won't look too difficult (as in the real GMAT),
then it's by far not as hard as it at first might look.

Even more, if you achieve to imagine what the two graphs look
like, it's quite trivial. In this case I intentionally chose those
"easy" relationship between the graphs, because the question
was designed to confuse most of you, but of course not all

By the way, 1:3 is correct.

Yes Sir. I totally agree that the solution is trivial if you reduce the equations to the known graphs..

BTW, There is a "Fun Math" section where such problems can be discussed. Here is the link...
http://www.gmatclub.com/phpbb/viewforum.php?f=38
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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