Cartesian plane

Hi

To graph in the cartesian plane
a --->$\displaystyle x^2+y^2 = 9$
would that be a graph where the x,y co-ordinates are (0,0) and the radius of the circle on the graph is 3 ???
(btw, while I know $\displaystyle 0^2+0^2 != 3^2$ the only example I have is an equation where $\displaystyle (x-10)^2 + y^2 = 16$, therefore (x,y) co-ords are (10,0) and radius is 4. From this I deduce the results from the equation above (a))

b ---> $\displaystyle x<=y$ has co-ords (x,y) = (0,0), which is a diagonal line through the intersection of x and y, from top right to bottom left, shaded area on upper side marking x<=y???

am i on the right track?

Seems you are wrong.

from $\displaystyle x^2+y^2 = 9$ we mean set of points (x,y) on the 2-d carteisian plane which satisfiy the given equation. (0,0) definately doesn't belong to this graph.

I'm just guessing,
but would this work?
$\displaystyle x^2 = 3^2 - y^2$
$\displaystyle x = 3 - y$

then x = 0 and y = 3

??

No, not even close

$\displaystyle (3^2-y^2)\neq (3-y)^2$

Well, i have no idea then
I would have thought
if x = 3, or -3
then y = 0

or if
y = 3 or -3
then
x = 0

because
$\displaystyle 3^2 + 0^2 = 9$ or any combination with 3, -3 and 0

so the points on the graph could be

(-3,0), (3,0), (0,-3), (0,3)

i wish I could draw it here, but basically it looks like a circle with center (0,0) and radius 3. (i am sure you can picture what I mean)

I don't know how else to approach it?

edit (sorry in my original post I said the x,y co-ords are (0,0) but now see this is the center and not the (x,y) co-ordinates)

yes - [tex]x^2+y^2 = 9[\math] is a circle with raidus = 3 and center = (0,0)

so all the points you said
(0,3),(0,-3),(3,0),(-3,0)
are on the circle and there are many more

But note (0,0) is not on that circle.

aman_cc and pickslides thank you very much.