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.
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samirpandeyit62
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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?
Eqn of circle with center at origin is x^2 + Y^2 =r^2
where r is the radius (x,y) any pt on the circle
(1)The radius of the circle is 4.
SUFF
(2)The sum of the coordinates of P is 0.
we cannot detemine the value from this as radius is not known.
A
what is the sum of the squares of the coordinates of P?
Eqn of circle with center at origin is x^2 + Y^2 =r^2
where r is the radius (x,y) any pt on the circle
(1)The radius of the circle is 4.
SUFF
(2)The sum of the coordinates of P is 0.
we cannot detemine the value from this as radius is not known.
A
Regards
Samir
Samir
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annakool1009
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-----------------samirpandeyit62 wrote: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?
Eqn of circle with center at origin is x^2 + Y^2 =r^2
where r is the radius (x,y) any pt on the circle
(1)The radius of the circle is 4.
SUFF
(2)The sum of the coordinates of P is 0.
we cannot detemine the value from this as radius is not known.
A
True that Equation of a circle centered at origin: x^2 + Y^2 =r^2
But, i doubt that (x,y) as above would be the co-ordinates any point P on the circle!
For ex: x =1, y = -1 (Sum of co-or = 0 and Sum of square of co-or = 2)
x1=2, y1= -2 (Sum of co-or = 0 and Sum of square of co-or = 8 )
Both (x,y) and (x1,y1) are on the circle and both satisfy the condition that the sum of the co-ordinates is '0', but the sum of the squares of the co-or is different and either of them [ (x,y) and (x1,y1) ] can be the co-ordinates of point P.
So shudnt statement A be infuff ???
Gearing up for the D-day.
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Morgoth
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The question in totality is asking for the square of the radius.annakool1009 wrote: True that Equation of a circle centered at origin: x^2 + Y^2 =r^2
But, i doubt that (x,y) as above would be the co-ordinates any point P on the circle
Equation of the circle
(x-h)^2 + (y-k)^2 = r^2
(h,k) are the cor-ordinates of the center of the circle.
Yes any point P on the circle would have the same distance from the origin but not the same co-ordinates. But the sum of squares of the co-ordinates would be same.
You can draw a circle on a xy plane and use pythagoreous theorem to find the co-ordinates.
For pythegeorous theorem, the standard four points would be (4,0) , (0,4), (-4,0) & (0,-4) now you can find any point P on the circle.
Hope its clear.
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kuroneko1313
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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?
Sorry I'm lost here; can anyone please explain what the question is asking for? What does it mean "the sum of the squares of the coordinates of P"? Should we know the coordinate of P first in order to answer this question? Please clarify!
-Elizabeth K.
what is the sum of the squares of the coordinates of P?
Sorry I'm lost here; can anyone please explain what the question is asking for? What does it mean "the sum of the squares of the coordinates of P"? Should we know the coordinate of P first in order to answer this question? Please clarify!
-Elizabeth K.
kuroneko1313 wrote: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?
what is the sum of the squares of the coordinates of P
let p be (x,y)
squares of the coordinates of P will be
x^2 and y^2
and sum will be
x^2 + y^2
as mentioned earlier
equation of circle for any point (x,y) on circle is
x^2 + y^2 = r^2
Sorry I'm lost here; can anyone please explain what the question is asking for? What does it mean "the sum of the squares of the coordinates of P"? Should we know the coordinate of P first in order to answer this question? Please clarify!
-Elizabeth K.
