Set of all points in a circle without distance formula The set of all points on a circle can be described by the equation $$x^2+y^2=r^2$$
How would you prove this, and is it possible to do so without the distance formula?
I ask because I want to know if the following proof of the Pythagorean theorem is circular:

Picture a circle with a point P somewhere on it. Draw a line from the center, O, to P. It will have a length equivalent to the radius $r$. The point P is located at $(x, y)$. We know the above equation, but if you think carefully, you'll notice that the line OP can be considered the hypotenuse of the triangle. The first base, lining up with the x-axis, is equal to $x$. The second, parallel to the y-axis, is equal to $y$. The second base is of course perpendicular to the first base. Therefore, given a right triangle with hypotenuse of length $r$ and bases of length $x$ and $y$, $x^2 + y^2 = r^2$.

(Proof rewritten from what is given in Apostol's calculus.)
Someone commented on the above

How is this not circular? How do you reasonably define a circle as the set of all points (x,y) such that $x^2 + y^2 = r^2$ etc without relying on the distance formula, which is an analytic version of the pythagorean theorem? Yes, I know it's customary to just define distance as the quantity you get from the distance formula, but if you picked that arbitrarily without relying on pythagorean theorem, there's no reason to believe that right triangles on the coordinate plane actually behave anything like right triangles in standard Euclidean geometry.

 A: 
The set of all points on a circle can be described by the equation
  x2+y2=r2
How would you prove this, and is it possible to do so without the
  distance formula?

====
Yes, both the circle formula $x^2 + y^2=r^2$ and the distance formula can both be proven from the Pythogorean Theorem.
Or you can prove the circle formula via the distance formula after proving the distance formula from the Pythogorean Theorem.
But you can not use the distance formula or the circle formula to prove the Pythogorean theorem.  That would be circular, because you have to use the Pythagorean theorem  to prove the distance formula in the first place.
(Unless the distance formula is given as an axiom.  But that would be very weird and difficult.  And intuitively it feels wrong.)
=====
So far as I can tell the following is not a proof of the pythagorean thereom, but a perfectly legitimate proving the formula of a circle from the pythagorean theorem.

Picture a circle with a point P somewhere on it. Draw a line from the
  center, O, to P. It will have a length equivalent to the radius r. The
  point P is located at (x,y). We know the above equation, 

I don't know what "the above equation" is.

but if you
  think carefully, you'll notice that the line OP can be considered the
  hypotenuse of the triangle. The first base, lining up with the x-axis,
  is equal to x. The second, parallel to the y-axis, is equal to y. The
  second base is of course perpendicular to the first base. Therefore,
  given a right triangle with hypotenuse of length r and bases of length
  x and y, x2+y2=r2.

This seems correct to me.  We take the Pythagorian thereom as a given.  We take and point on a circle.  We make a right triangle out of it.  The sides of the triangle are $x, y$ and hypotenuse $r$. So $x^2 + y^2 =r^2$ is the formula.
This is perfectly correct.  
To be thorough we actually need to prove that if $(x,y)$ in a point in so that $x^2 + y^2 = r^2$ then that point must lie on the circle.  The prover left that out.  But it's the exact same argument in reverse. The point forms a right triangle.  And the hypotenuse must be $r$ so $(x,y)$ is on the circle.
So now the complaint.

How is this not circular? How do you reasonably define a circle as the
  set of all points (x,y) such that x2+y2=r2 etc without relying on the
  distance formula, which is an analytic version of the pythagorean
  theorem?

This would be a very legitimate and correct complaint... except so for as I can see this doesn't apply to the original post at all!  The original post did not define the circle as points $(x,y)$ such that $x^2+y^2= r^2$ but actually proved that from the definition a circles radius is constant.
So I don't get his/her complaint.  I suspect s/he is actually writing about another poster????
But his/her comment:

Yes, I know it's customary to just define distance as the quantity you
  get from the distance formula, but if you picked that arbitrarily
  without relying on pythagorean theorem, there's no reason to believe
  that right triangles on the coordinate plane actually behave anything
  like right triangles in standard Euclidean geometry.

Is absolutely spot on!  It just doesn't apply to any error the original poster made.
===
Anyway:
We can (have to) take Euclid's 5th postulate as an axiom.
We use Euclid's 5th postulate to prove pythogorean theorem.
We can just as easily take the pythogorean theorem as an axiom and prove Euclid's 5th postulate.  We traditionally don't do that as Pythogorean Theorem without justification is not intuitively obvious whereas Euclid's 5th postulate.
It's usually assumed that before a student starts analytic geometry, that basic geometry of Euclid's 5th postulate and the pythagoren theorem is a given.
However.  Euclid's 5th postulate in analytic terms is simply the axioms that lines are of the equation $y = mx + b$ where $m$ is a constant slope.
We can prove Pythogorean Theorem analytically by... well if I draw a square $(0,0), (0, a+b), (a+b,a+b), (a+b, 0)$ that square will have arean $(a+b)^2 = a^2 + 2ab + c^2$.  (We can see that visually by drawing the line $(0,a)$ to $(a+b, a)$ and the line $(a,0) $ to $(a, a+b)$.  That cuts the square into $4$ regions; a square with side $a$ and area $a^2$, a square of side $b$ and area $b^2$ and two rectangles with sides $a$ and $b$ and area of $ab$.)
We draw the lines $(0, b)$ to $(a,0)$ and $(a,0)$ to $(a+b, a)$ and $(a+b, a)$ to $(b, a+b)$ and from $(b, a+b)$ to $(0,b)$,  This breaks the exact same square into $4$ right triangles with sides $a,b$ and hypotenuse $c$ and a square with sides $c$.  So the area of the square is both $4*\frac 12 ab + c^2$ and $a^2 + 2ab + b^2$.
So $a^2 + b^2 = c^2$.
From here we can either prove that the formula for a circle is $x^2 + y^2 =r^2$ as was done above.  Or prove the distance formula by a nearly identical argument.
Now, hypothetically, I suppose one could take either the circle formula or the distance formula as an axiom an prove that lines are of the form $y = mx +b$.  But that'd be really weird and there really isn't much point.
A: A circle is defined as
the set of all points
at the same distance from
a point.
If the distance is $r$
and the point is
$(a, b)$,
then, from the
Pythagorean theorem,
each point
$(x, y)$
satisfying
$(x-a)^2+(y-b)^2 = r^2$
is distance $r$
from $(a, b)$
and each point
of distance $r$ from
$(a, b)$
satisfies
$(x-a)^2+(y-b)^2 = r^2$.
Also,
any point that does not satisfy
$(x-a)^2+(y-b)^2 = r^2$
is not distance $r$
from $(a, b)$.
Is this better?
