Coordinate

[N:Dure]

??

[fskilnik@GMATH]

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.)

[N:Dure]

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?

[GMATGuruNY]

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.

[N:Dure]

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)

[GMATGuruNY]

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!

[fskilnik@GMATH]

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!