Is a circle still considered a surface?

HAL10000

The question asks to look for a surface and a circle is the only function which meets the conditions. Is this still considered a surface?

DonAntonio

If it is really true that only the circle meets the conditions then there's hardly anything to talk about here, but the

interesting thing is, imo: what exactly is your definition of surface?

HallsofIvy

No, a circle is a surface if and only if you are talking about the "general" "n dimensional surface" in which case you can think of a circle as a "1 dimensional surface". Of course, it would help a lot if you told what the "question" really is!

HAL10000

The question asked for a surface which is equidistant from all points p(x,y,z) to the point (0,0,1) and the plane through z=-1

Mark44

I don't see that a circle figures into this problem at all. In the plane, a parabola is equidistant from a given fixed point and a given line. In other words, at each point P on the parabola, the distance from P to the fixed point is equal to the distance from P to the line.

theorem4.5.9

Although a circle is a perfectly good 1 dimensional surface, it's not the solution to your problem.

I believe your problem is asking for this:
Find all points ##(x,y,z)\in \mathbb{R}^3## such that
$$\mathrm{distance}\left( (x,y,z) , (0,0,1) \right) = \mathrm{distance}\left( (x,y,z) , \mathrm{plane} \right) $$

I have written it this way as to not give away the answer.

HAL10000

Cool :) and I was so certain about that circle lol. Thanks a lot. I get a circular prabolloid with a and b = 2.