
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 coordinates 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 $\displaystyle (x10)^2 + y^2 = 16$, therefore (x,y) coords are (10,0) and radius is 4. From this I deduce the results from the equation above (a))
b > $\displaystyle x<=y$ has coords (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 2d 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
$\displaystyle (3^2y^2)\neq (3y)^2$

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 coords are (0,0) but now see this is the center and not the (x,y) coordinates)

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.