To graph in the cartesian plane
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 the only example I have is an equation where , therefore (x,y) co-ords are (10,0) and radius is 4. From this I deduce the results from the equation above (a))
b ---> 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?
July 20th 2010, 09:47 PM
Seems you are wrong.
from 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.
July 20th 2010, 10:08 PM
I'm just guessing,
but would this work?
then x = 0 and y = 3
July 20th 2010, 10:12 PM
No, not even close
July 20th 2010, 10:24 PM
Well, i have no idea then
I would have thought
if x = 3, or -3
then y = 0
y = 3 or -3
x = 0
because 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)
July 20th 2010, 11:05 PM
yes - [tex]x^2+y^2 = 9[\math] is a circle with raidus = 3 and center = (0,0)
so all the points you said
are on the circle and there are many more