1
$\begingroup$

I want to plot the equation $x^2+y^2=z$ inside a volume. However I don't want to plot the center of the graph on the origin. "inspired" by the equation of a circle I tried: $(x-2)^2+(y-2)^2$ with hope of plotting the lowest point of the graph on the point $(2,2,0)$. I then tried $x+y=z$ and $(x-2)+(y-2)=z$ but this only affected it by $4$ on the $x$ axis, because it simplifies to $x+y-4=z$. (though I don't understand why this affects the $x$ axis and not the $y$ or the $z$ axi?)

How can I translate either Of those equations?

1 Answers 1

1

So there are a couple of things going on here. Firstly, you are correct in one sense. $x^2 + y^2 + z^2= 4$ is the equation for a sphere centered around the origin in 3-space. $(x-2)^2 + (y-2)^2 + z^2 = 4$ is that same sphere, but with the center at $(2,2,0)$. So there, you are correct.

But in your second, $x + y - 4 = z$ or rather $x + y - z = 4$, these are planes in 3-space. The four offsets all three, x, y, and z coordinates (not just x). For example, one can see where the plane intercepts the three axes. In the case without the 4, all intercepts are at 0. But now, the x intercept is 4, the y intercept is 4, and the z intercept is -4. You might think - that's really weird! But a plane is rigid, and moving the x part affects the y part too.

Does that all make sense?

  • 0
    @Jonathan (for reference, we see that both intercepts are 4 in the second graph, which we expected).2011-09-05