For a circle with radius of $5$, we know it is,
$$x^2 + y^2 = 25.$$
I can interpret it as $f(x,y) = x^2+y^2$ and $g(z)=25$, the final circle is when the two graph intersects with each other in 3D space.
But how to interpret it when I transform it into this,
$$x^2 = 25 - y^2.$$
$f(x) = x^2$ is parabola in 2D space, $g(y) = 25 - y^2$ is another parabola in 2D space, the two parabola intersection will never be a circle, I know there is something wrong goes here, but I want to plot the two functions to see the circle shape.
Edit:After plotting the graph, I've got this
And if take a bird view, it shows the circle projected on xy plane
What I done wrong is I tried to draw $x^2$ part on the xy plane as I would normally do with y=$x^2$, but it seems I should use the Z variable to draw the graph z=$x^2$ instead of y=$x^2$.
REBOOT (you can see my earlier posts in the edit).
The intersection of $z = x^2 + y^2$ and $z=25$ is $x^2 + y^2 = z =25$. That is a circle floating $25$ feet in the air.
The intersection of $z = x^2$ and $z= 25-y^2$ is $z = x^2 = 25-y^2$ or $x^2 + y^2 = 25 = z+y^2$. That is a horse on a merry-go-round. The horse's pole goes around in a circle while the horse itself goes up and down.
===== tl;dr =======
REBOOT (you can see my earlier posts in the edit).
If we consider a curve in $2$D space as $^1C = \{(x,y)|$ some relation between $x,y\}$ this exists entirely in the $x,y$ plane.
If we consider this as somehow related to curve $^2C_1$ in $3$D space where $^2C_1 = \{(x,y,z)|$ the same relationship between $x,y$ but also a $3$rd dimension $z$ is related somehow$\}$ then the we are considering $^1C$ and a projection of $^2C_1$ from the $x,y,z$ $3$D space down into $^1C$ in the $x,y$ $2$ D plane. via projection $\iota: (x,y,z) = (x,y)$ and $^1C = \iota(^2C_1)$.
This is fine. The $z$ value is ignored.
So if we consider $^2C_1$ is the intersection of the parabloid graph of $f(x,y) = x^2 + x^2$ and planar graph $g(x) = 25$ the intersection of a parabloid and a plane perpendicular to the axis of symmetry of the parabloid is a circle.
And indeed we get $^2C_1= \{(x,y,z)|x^2 + y^2 = z = 25\} =\{(x,y, 25)| x^2 + y^2= 25\}$; a circle floating $25$ feet in the air. And if we project it down to the $x,y$ plane get $\iota ^2C_1 = \{(x,y)| x^2 + y^2=25$ and we completely ignore $z\}=S=\{(x,y)| x^2 + y^2 = 25\}$; a circle... the exact same circle in every way but we've pushed it down onto the ground.
Projecting a curve from a plane $z=25$ to the $x,y$ plane (which if we viewed it as part of 3D space would be $z=0$) is trivial an we do it without thinking; the $z$ doesn't matter, it could be $0$ or $25$ or $-\sqrt \pi$ for all we care, we're just shifting the curve up or down.
But a projection is ignoring the $z$ value, which is different than assuming it is a constant.
If $^2C_2 = \{(x,y,z)|$ some relation between $x,y,$ and $z$ and $z$ can go up and down then $\iota ^2C_2 =\{(x,y)|$ the same relation between $x,y$ but the $z$ value is ignored$\}$ is a projection shadow information is lost.
So if we consider you two functions in $3$D space.
$f(x,y) = x^2$ and $g(x,y)=25 -y^2$. The graph of $f(x,y)$ is parabola shaped valley that cuts through $3$D space. It's a parabola that extends into the third dimension along the $y$ axis. And $g(x)$ is the same thing, but upside down and extending along the $x$ axis.
We can intersect them and the curve will be $^2C_2=\{(x,y,z)| z= x^2 =25 - y^2\}$. What does this curve look like. It's probably not a circle. And if we manipulate symbols $^2C_2 = \{(x,y,z)| x^2 + y^2 = 25 = z+y^2\}$. That is is .... well it's the path of a person riding on a merry-go-round. $x^2 +y^2 = 25$ so the horsey-pole goes around in circle, but the horse goes up and down from $y=0$ and $z =25$ to $y=5$ and $z= 0$.
Okay.... so what does this look like when we ignore the $z$ up-down motion and project it into the $x,y$ plane.
Well, it is $\iota ^2C_2 = \{(x,y)|x^2 + y^2 = 25$ and we completely ignore that $z = 25 + y^2\}=\{(x,y)|x^2 +y^2 =25\} = S$. That's the exact same circle.