Is a < 2?

(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3. This circle is centered at the origin and has the radius of \(\sqrt{3}\approx{1.7}\) (check here: math-coordinate-geometry-87652.html). No point inside that circle has x-coordinate greater than \(\sqrt{3}\approx{1.7}\), thus \(a<1.7<2\). Sufficient.

(2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10 --> 2*4+4x=10 --> x=1/2. Sufficient.

Answer: D.
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Hi All,

As complex as this graphing-based DS question looks, the given Facts focus on two equations, so they're relatively easy to deal with/solve.

We're asked if A is less than 2. This is a YES/NO question.

Fact 1: In an XY plane, point (A,1) lies inside the circle whose equation is X^2 + Y^2 = 3

With the given equation and the given (X,Y) co-ordinate (A,1), we can plug in and solve for A....

A^2 + 1^2 = 3
A^2 = 2

Since we're dealing with a circle, there are two possible points that have a Y-co-ordinate of 1: (A, 1) and (-A, 1).

Since A^2 = 2, A = Root(2).

Root(2) is < Root(4)

Root(2) is < 2, so whether we're dealing with A = Root(2) or A = -Root(2), the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT.

Fact 2: In an XY plane, point (A,4) lies on the line whose equation is 2Y + 4X = 10

Here, we have a simple line to work with, so we can substitute in the value of A (for X) and 4 (for Y), giving us....

2(4) + 4(A) = 10
8 + 4A = 10
4A = 2
A = 1/2
The answer to the question is YES.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi Rich!
Nice solution as usual! But please correct a typo. The bold part should be "4A" not A=))
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Hi Konstantin1983,

Good eye. I've made the correction. Thanks for catching it.

GMAT assassins aren't born, they're made,
Rich
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Ah Rich no problem=)). We are human being and even assassins sometimes make typos=))
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1. We are told that a is the x coordinate for a point within a circle with radius sqrt(3) its center is origo.

The largest x coordinate possible would be if y=0, otherwise x would have to decrease(if +) or increase (if -) to make sure the distance from origo does not exceed sqrt(3).
If y=0, a can be at most sqrt(3). This is less than 2. As y is 1, a has to be even smaller.
Sufficient.

2.
We can solve for x. Sufficient.

Answer D

Posted from my mobile device
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Bunuel: Was able to understand your solution, have a query as to how the circle is centered at the origin. Is it a default assumption or its inferred from the equation.?

Equation of the circle is given by the formula,

\((x-a)^2 + (y-b)^2 = r^2\), where a,b are the center coordinates and r is the radius.

Now, in the question we are given the form

\((x)^2 + (y)^2 = r^2\), implying a and b are zero. Or the circle is centered at origin.

I think you should go through the basic of Circle once. That would help you in such questions.
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Bunuel - Statement 1 says point lies inside the circle so I get a = + and - 2 as 2 values I can confirm that since point is inside the circle then it has to be less than 2 and greater than -2 [i.e. -2 <a<2 ] does that make sense ?

\(a<2\)?

Looking at S-2 because it seems simpler:

2) If (a,4) lie on the line \(2y + 4x = 10\), then \(2(4) +4x=10\) ---> \(x=\frac{1}{2}\). Sufficient

1) If (a,1) lies inside the circle defined by \(x^2 + y^2 = 3\), a simple way to check whether a<2 at this point on the x,y plane, would be to sub value of y and check value of x. \(x^2 + 1^2= 3\)---> \(x^2=2\)---> \(x=\sqrt{2}<2\). If (a,1) lies within the circle and value of \(x<2\) when \(y=1\) this definitively means that \(a<2\)Sufficient.

Statement 1:
Assume the point lies on the circle, we can plug in the point to get \(a^2 + 1^2 = 4\), or \(a^2 = 3\). Then the points \((-\sqrt{3}, 1)\)and \((\sqrt{3}, 1)\) are on the circle. For the point to be within the circle, we must have roughly\(-1.7 < a < 1.7\), therefore \(a < 2\) is true. Sufficient.

Statement 2:
We are allowed to solve for the coordinates of \((a, 4)\)with this line, sufficient.
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If you don't have the equation of a circle memorized, can't you just plug in (a,1) into x^2 + y^2 = 3 to prove statement I sufficient?
--> a^2 = 2 --> a = + or - sqrt(2) Both the positive and the negative solution are less than 2, so statement I is sufficient.

In statement-1 how do we know that the centre of the circle lies on the orign?

Hi sarthakaggarwal,

The easiest way for you to prove that the graph is centered around the Origin (0,0) might be to simply graph it for yourself. Mathematically-speaking though,
with the equation X^2 + Y^2 = 3, since both the X and the Y are the squared terms, the Original will always be the center of the circle (regardless of how big the circle is. However, if you change either of the squared terms (for example... (X+1)^2 + (Y - 2)^2 = 3.... then the center of the circle will shift - and in this case, the center would be at (-1,+2).

GMAT assassins aren't born, they're made,
Rich
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HI EMPOWERgmatRichC, thanks for the reply
I know that (x-h)^2 + (y-k)= r^2 and if both H and K are equal to 0 then the centre lies on the Origin.
but when the equation was given as X^2 + Y^2 = 3. I thought that some value of "H"and some value of "K" has already been subtracted from "X" and "Y" respectively.
So basically, if in the question the equation of a circle appears as X^2+Y^2=r^2 can i assume it that the centre of the circle lies on the origin?
Thanks in advance

Hi sarthakaggarwal,

YES - as long as the 'pieces' of that equation are written as "X^2" and "Y^2", you know that the center of the circle is at the Origin.

GMAT assassins aren't born, they're made,
Rich
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