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Posts: 23
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A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?
1) The radius of the circle is 4.
2) The sum of the coordinates of P is 0.
1) The radius of the circle is 4.
2) The sum of the coordinates of P is 0.
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Sallyzodiac
Say the coordinates of P are (x,y), then the question asks about the value of x^2 + y^2.
In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)
This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.
If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\).
According to the above, the first statement of the question gives the direct answer: x^2 + y^2 = r^2 = 4^2. The second statement is not sufficient.
Check more here: math-coordinate-geometry-87652.html
Hope it helps.
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Updated on: 21 Jun 2016, 07:10
Originally posted by FacelessMan on 03 May 2016, 08:40.
Last edited by FacelessMan on 21 Jun 2016, 07:10, edited 1 time in total.
Last edited by FacelessMan on 21 Jun 2016, 07:10, edited 1 time in total.
Hi Sallyzodiac,
The center-radius form of the circle equation is in the format \((x – h)^2\) + \((y – k)^2\) = \(r^2\), with the center being at the point (h, k) and the radius being "r". Now if we are told that the circle lies on the origin, this equation is reduced to \(x^2 + y^2 = r ^2\).
So, for us to calculate the squares of the coordinates - we just need to know the radius of the circle. Hence A.
The center-radius form of the circle equation is in the format \((x – h)^2\) + \((y – k)^2\) = \(r^2\), with the center being at the point (h, k) and the radius being "r". Now if we are told that the circle lies on the origin, this equation is reduced to \(x^2 + y^2 = r ^2\).
So, for us to calculate the squares of the coordinates - we just need to know the radius of the circle. Hence A.
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General Discussion
Posts: 23
Could someone please elaborate what the term "sum of the squares" mean in the context of coordinate geometry?
Hi Bunuel - My confusion is on the wording of the question- " If p is a point on the circle..." this is in fact saying that p has to be on the "outlined" circle and NOT for example inside the circle? If we choose p to be in the origin, 0,0 we would have another answer. But from what I am reading no one is considering P to be "inside" the circle. Are there any key clue to rule out points inside the circle?
best
Oloman
best
Oloman
Bunuel
oloman
Yes, on the circle means on the circumference, while in the circle means inside the circumference.
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Question asked here is, nothing the square of the radius of the circle, since the circle is centered at origin. Statement that helps in finding out the radius of the circle is sufficient.
Hence A.
Hence A.
Keats
We are given the center at origin, so equation of circle will be \(x^2 + y^2 = r^2\)
Since, P is on the circle, it will satisfy this equation.
Statement 1 : We are given the value of r = 4, so we can find out the value of \(x^2 + y^2\), Hence Sufficient.
Statement 2 says x = -y. Even if we put this value in the equation we will get \(2x^2 = r^2\)
But Since we don't know r, we cannot find the value of x. hence, insufficient.
Answer A.
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Posts: 16
My god... Back in High School (or maybe elementary school), here in Portugal, we distinguish a circumference from a circle as being geometry elements completely different from each other. While a circumference is an imaginary line in which every point has the same distance to a specified center, defining thus the limits of a certain circular area and the circle is that area and is defined by the infinite set of points that are in a distance less than or equal to a specified radius. So, a circumference is and equation such as x^2 + y^2 = r^2 and a circle is an inequality such as x^2 + y^2 <= r^2. So the circumference has distance units such as meter, miles, feet, etc and the circle has square distance units such as square meters, etc. This is brilliantly coherent and makes no room for any confusion!
As far as I can understand on the GMAT, according to Bunnel:
"Yes, on the circle means on the circumference, while in the circle means inside the circumference. "
This distinction is SUPER confusing!!! in/on makes all the difference and while for a native this might be even logic for a non native is a pain in.... Furthermore, this makes a person who knows all the mathematical concepts to answer this question right, answering it wrong because of english language issues, which are not supposed to be assessed in this part...
I'm sorry for unburdening my frustration on you guys but I got 2 questions wrong on the same mock because of this and my exam is in 5 freaking days...
Cheers to y'all!
As far as I can understand on the GMAT, according to Bunnel:
"Yes, on the circle means on the circumference, while in the circle means inside the circumference. "
This distinction is SUPER confusing!!! in/on makes all the difference and while for a native this might be even logic for a non native is a pain in.... Furthermore, this makes a person who knows all the mathematical concepts to answer this question right, answering it wrong because of english language issues, which are not supposed to be assessed in this part...
I'm sorry for unburdening my frustration on you guys but I got 2 questions wrong on the same mock because of this and my exam is in 5 freaking days...
Cheers to y'all!
Posts: 1,142
Miracles86
Hello
May I suggest you dont worry about it too much. You are correct that a circle centered at origin (0,0) with a radius of r should ideally be written as:
x^2 + y^2 <= r^2
And I think the meaning of this inequality is that within the circle (inside, not touching the boundary), all points will satisfy: x^2 + y^2 < r^2
While on the boundary (circumference), all points will satisfy: x^2 + y^2 = r^2
Now as per the first statement, since radius of the circle is 4, and since P is a point (x,y) on the boundary of the circle, then definitely the sum of squares of the two coordinates of P, x^2 + y^2 will be = r^2. First statement is thus sufficient.
Posts: 16
amanvermagmat
Of course, that is absolutely right. But when I read the question, I thought that this was a trap for rushing test takers. So I thought the point was somewhere inside the circle and because it wasn't on the boundary, as you properly deduced, no conclusion could be made about the squares of x and y... Got it now and at least this mistake I won't make it again!
Thanks for your tips!
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Guys, the easiest way to remember the formula and information pertaining to the properties of this question is to recall that the radius is found in the same way the distance between two points is found.
If you remember that x^2 + y^2 = r^2 Then remember that any point on the circumference of a circle in the xy plane will have the SAME distance from the origin. This distance is given by the radius. You can back solve to clarify your understanding.
If you remember that x^2 + y^2 = r^2 Then remember that any point on the circumference of a circle in the xy plane will have the SAME distance from the origin. This distance is given by the radius. You can back solve to clarify your understanding.
Posts: 1
the question must have said "on the circumference" rather than "on the circle"
SalmonSteak
Logically speaking, a point cannot lie on the circumference. Circumference is the measure of something and a point cannot lie on the measure of something. on the circle implies that the point lies on the boundary/border.
Also, Bunuel - this is a gmatprep question.
Brian123
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Added the tag. Thank you!
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A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?
The Equation of a circle with center (h,k) and radius r is given by
\((x-h)^2 + (y-k)^2 = r^2\)
If the center of the circle is at the orgin (0,0) , we can reframe the equation as
\(x^2 + y^2 = r^2\)
If P is a point on the circle i.e \((x,y)\), we are asked to find the sum of the squares of the coordinates of P. i.e. \(x^2 + y^2\)
We already know that \(x^2 + y^2 = r^2\), so in order to get the answer for the question stem, we need to find the radius of the circle.
(1)The radius of the circle is 4.
Since the radius of the circle is given, \(x^2 + y^2 = 4^2\). Hence It's sufficient.
(2)The sum of the coordinates of P is 0.
It says that coordinates of point P could be \((x,-x)\) or \((-x,x)\) where in both cases the sum is 0.
Substituting these values in the equation \(x^2 + y^2 = r^2\)
\(x^2 + (-x)^2 = r^2\)
\(2x^2 = r^2\)
Since we dont have the value for r or x, Statement 2 alone is insufficient.
Option A is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
The Equation of a circle with center (h,k) and radius r is given by
\((x-h)^2 + (y-k)^2 = r^2\)
If the center of the circle is at the orgin (0,0) , we can reframe the equation as
\(x^2 + y^2 = r^2\)
If P is a point on the circle i.e \((x,y)\), we are asked to find the sum of the squares of the coordinates of P. i.e. \(x^2 + y^2\)
We already know that \(x^2 + y^2 = r^2\), so in order to get the answer for the question stem, we need to find the radius of the circle.
(1)The radius of the circle is 4.
Since the radius of the circle is given, \(x^2 + y^2 = 4^2\). Hence It's sufficient.
(2)The sum of the coordinates of P is 0.
It says that coordinates of point P could be \((x,-x)\) or \((-x,x)\) where in both cases the sum is 0.
Substituting these values in the equation \(x^2 + y^2 = r^2\)
\(x^2 + (-x)^2 = r^2\)
\(2x^2 = r^2\)
Since we dont have the value for r or x, Statement 2 alone is insufficient.
Option A is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
IanStewart
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The equation of a circle centred at the origin is x^2 + y^2 = r^2, where r is the radius of the circle. So Statement 1 is sufficient; the sum of the squares of the coordinates will be 4^2 = 16. Statement 2 only tells us that for point P, y = -x, so point P is on the line y = -x. But that tells us nothing about the circle, because any circle big or small that is centred at the origin will meet the line y = -x in two places, so the answer is A.
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IanStewart
Hi Ian,
Could you elaborate a bit more on statement 2? For e.g. if the point is in the 1st coordinate, I would think that both x and y coordinates should be positive, so I wonder where do we get the x = -y equation from?
Thanks
Hi Ian,
Could you elaborate a bit more on statement 2? For e.g. if the point is in the 1st coordinate, I would think that both x and y coordinates should be positive, so I wonder where do we get the x = -y equation from?
Thanks
TargetMBA007
(2) states that the sum of the coordinates of P(x, y) is 0. Therefore, x + y = 0, which leads to y = -x. This indicates that point P lies on the line y = -x.
Note that any point on this line will satisfy x + y = 0, and P is positioned in either the II or IV quadrants.
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