its homework and i dont know the answer please help
and what equation does it represent if it isn't a circle ?
and what equation does it represent if it isn't a circle ?
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General circle form: x^2 + y^2 = (radius)^2
Well, if you have radius^2 = 0, radius = 0
You have a circle with radius of 0... giving you one point at (0,0)
The definition of a circle is the set of points in a plane equidistant from a central point.
I suppose one point is equidistant from itself, so one could argue that a circle with the radius of 0, a singular point, is teeeechnically a circle, but on the other hand it isn't at the same time.
It's just one of those iffy technicalities... Meh. I wouldn't worry about that so much, just recognize that x^2+y^2=0 represents a single point
Well, if you have radius^2 = 0, radius = 0
You have a circle with radius of 0... giving you one point at (0,0)
The definition of a circle is the set of points in a plane equidistant from a central point.
I suppose one point is equidistant from itself, so one could argue that a circle with the radius of 0, a singular point, is teeeechnically a circle, but on the other hand it isn't at the same time.
It's just one of those iffy technicalities... Meh. I wouldn't worry about that so much, just recognize that x^2+y^2=0 represents a single point
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x^2+y^2=r^2
This is the equation of a circle with radius "r".
In the equation you provided, r=0. In other words, we have a circle with no radius, which means we don't have a circle at all.
x^2+y^2=0 is only defined at one point.
Which makes sense, because if you try to solve for y, you see that:
y^2 = -x^2
y = +- sqrt(-x^2)
But the domain of a square root function is only defined for values that are non-negative. And no matter what x-values we try to plug in, only ONE x-value gives a non-complex answer.
When x=0, y=0.... Since (0,0) is the only thing defined for this equation, the equation is simply a point.
This is the equation of a circle with radius "r".
In the equation you provided, r=0. In other words, we have a circle with no radius, which means we don't have a circle at all.
x^2+y^2=0 is only defined at one point.
Which makes sense, because if you try to solve for y, you see that:
y^2 = -x^2
y = +- sqrt(-x^2)
But the domain of a square root function is only defined for values that are non-negative. And no matter what x-values we try to plug in, only ONE x-value gives a non-complex answer.
When x=0, y=0.... Since (0,0) is the only thing defined for this equation, the equation is simply a point.
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The right side of the equation represents the radius squared.
For example:
x^2 + y^2 = 4
then the radius is 2 because 2^2 is 4
Since the right side of your equation is 0, the square root of 0 is 0. This means the radius of your equation would be 0 but if the radius is 0, then the circle has no radius and it doesn't exist.
To answer your second question, this equation doesn't exist.
For example:
x^2 + y^2 = 4
then the radius is 2 because 2^2 is 4
Since the right side of your equation is 0, the square root of 0 is 0. This means the radius of your equation would be 0 but if the radius is 0, then the circle has no radius and it doesn't exist.
To answer your second question, this equation doesn't exist.
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General formula for a circle centered at the origin is x^2 + y^2 = r^2 where r is the circle's radius. However, because in this situation r^2 = 0, that means the circle's radius is 0. If there's no radius, then there's no circle. It's one of those conceptual trick questions lol
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The part to the right of the = sign is the radius^2, so the radius would be 0, or in other words not a circle at all.