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EMPOWERgmatRichC
Hi All,


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

Final Answer:
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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.

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Hi Konstantin1983,

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

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Rich
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

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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.?
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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
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: https://gmatclub.com/forum/math-coordina ... 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.

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 ?
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Is a < 2?

(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3
(2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10

\(a<2\)?

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.
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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.
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In statement-1 how do we know that the centre of the circle lies on the orign?
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sarthakaggarwal
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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sarthakaggarwal
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
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
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sarthakaggarwal
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
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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ranjitaarao
Is a < 2?

(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3
(2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10

Hey Bunuel

This is how I'm thinking of statement 1

I know (x-a)^2+(y-b)^2<r^2 when point (a,b) lie inside the circle. Similarly, I know - (1)
(x-a)^2+(y-b)^2>(r^2) when point is outside the circle
(x-a)^2+(y-b)^2=(r^2) when points lie inside the circle.

inputting values of (a,1) in (1) , i get a^2<2 & hence value of a lies -√ 2<a<√ 2 . Hence, a is definitely less than 2.

Statement 2
Gives me a=1/2 after inputting the values of (a,4) in the equation of the line.

Does this method work?
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