So I have a problem figuring out how to find the coordinates and radius to quadratic equations that are not in the form of $$(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 $$
Where the coordinates are going to be: $$(x_0,y_0,z_0)$$
But when I get something like: $$ \left\{\begin{matrix}x^2 + y^2 + z^2 = 4 \\ x + y + z=1 \end{matrix}\right. $$
Now the solution manual says this is centered at $$ ( \frac{1}{3}, \frac{1}{3}, \frac{1}{3}) $$
With the radius at: $$ \sqrt{4- \frac{3}{9}} = \sqrt{\frac{11}{3}} $$
Now I don't see any connection to how one would come to that conclusion with the given information. The book says nothing about it. It say that this is a sphere and a plane that intersect each other.
The next problem is the same idea: $$ \left\{\begin{matrix}x^2 + y^2 = 1 \\ z = x + y \end{matrix}\right. $$ This object is an ellipse intersecting a cylinder, with the coordinates: $$ +- (1\sqrt{2}, 1\sqrt{2},\sqrt{2}) $$ To me this makes no sense at all. I would love for someone to teach me how to get these coordinates and radius with the given parametric information. Cheers!
Preview
1. System of equations $$ \left\{\begin{matrix}x^2 + y^2 + z^2 = 4 \\ x + y + z=1 \end{matrix}\right. $$ The first equation is a sphere around the origin $0$ with radius $r = 2$. The second equation is a plane $E$ through $e_1 = (1,0,0)^T$, $e_2 = (0,1,0)^T$ and $e_3 = (0,0,1)^T$.
If one combines both equations, one gets the equations of the intersection curve.
We now use the plane equation to express $z$ as function of the $x$ and $y$ coordinates: $$ z = 1 - x - y \quad (\#) $$
Putting this into the first equation gives: \begin{align} 4 &= x^2 + y^2 + (1 - x - y)^2 \\ &= x^2 + y^2 + 1 + (x + y)^2 - 2(x + y) \\ &= x^2 + y^2 + 1 + x^2 + y^2 + 2xy - 2(x+y) \\ &= 2(x^2 + y^2 + xy -x -y) + 1 \end{align} which gives a conic section $A x^2 + B xy + C y^2 + D x + E y + F = 0$: $$ 3/2 = x^2 + y^2 + xy - x - y \quad (*) $$ The discrimintant $\Delta = B^2 - 4AC = 1 - 4 = -3 < 0$ is negative, so it is an ellipse $C_{xy}$ in the $x$-$y$-plane.
Note: The intersection curve $C$ we get via $(\#)$: $$ C = \{ (x,y,1-x-y)^T : (x,y)^T \in C_{xy} \} $$ Back to analyzing $C_{xy}$:
The special case of a circle would require $A = C$ and $B = 0$. So it is no circle.
Equation $(*)$ can be written as: $$ u^T A u + b^T u = 1 $$ with $$ A = \frac{2}{3} \left[ \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) + \left( \begin{matrix} 0 & 1/2 \\ 1/2 & 0 \end{matrix} \right) \right] = \frac{2}{3} \left( \begin{matrix} 1 & 1/2 \\ 1/2 & 1 \end{matrix} \right) = \left( \begin{matrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{matrix} \right) $$ and $$ b = \frac{2}{3} \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -2/3 \\ -2/3 \end{matrix} \right) $$ $A$ has the eigenvectors $(1,1)^T$ and $(1, -1)^T$ with the positive eigenvalues $1$ and $1/3$. From this one can infer it is an ellipse as well.
Transforming to normed eigenvectors via $v = P u$ with $$ P = \left( \begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix} \right) \quad P^{-1} = P = P^T $$ we get $$ v^T P A P v + (Pb)^T v = 1 $$ where $PAP = \mbox{diag}(1, 1/3)$ and $(P b)^T = (-\sqrt{8}/3, 0)$. Using $v = (\xi, \eta)^T$ we get $$ \xi^2 + 1/3 \eta^2 - \sqrt{8}/3 \xi = 1 \iff \\ (\xi - \sqrt{8}/6)^2 + 1/3 \eta^2 = 1 + 8/36 = 1 + 2/9 = 11/9 \iff \\ (\xi - \sqrt{2}/3)^2 + (\eta / \sqrt{3})^2 = (\sqrt{11}/3)^2 \iff \\ \left(\frac{\xi - \sqrt{2}/3}{\sqrt{11}/3}\right)^2 + \left(\frac{\eta}{\sqrt{11/3}} \right)^2 = 1 \quad (**) $$ This is an equation of the form $$ \left(\frac{x - x_0}{a}\right)^2 + \left(\frac{y-y_0}{b}\right)^2 = 1 $$ So in the $\xi$-$\eta$-coordinate system, which is mirrored along the $x$-axis and then rotated $-45^\circ$, the projection $C_{xy}$ is an ellipse with origin $v_0 = (\xi_0, \eta_0) = (\sqrt{2}/3, 0)$, semi-axis $a = \sqrt{11}/3 \approx 1.11$ in $\xi$-direction, and semi-axis $b = \sqrt{11/3}\approx 1.73$ in $\eta$-direction.
Transforming back we have $u_0 = P v_0 = (1/3, 1/3)^T$ as center in the $x$-$y$-plane. This gives a center $U_0 = (1/3, 1/3, 1-1/3 -1/3)^T = (1/3, 1/3, 1/3)^T$ of $C$.
Took a bit to reach the first statement of your instruction manual. :-)
Next bit is determing $C$ from $C_{xy}$.
One way is to parameterize $C_{xy}$ in the $\xi$-$\eta$-system as $v(t)$ then transform back to $u(t) = P v(t)$. Then go into the third dimension via $U(t) = g(u(t))$. Then go into local coordinates into the plane $w(t) = f(U(t))$.
The expectation is that this is a circle around $w_0 = (0,0)^T$ with radius $\sqrt{11}/3$, the same value as the major semi-axis.
2. Here we have the system $$ \left\{ \begin{matrix} x^2 + y^2 = 1 \\ z = x + y \end{matrix} \right. $$
This system is solved by the intersection curve of a cylinder and a plane.
From the above image we suspect it is some ellipse as well.
$$ u = (x, y, z)^T = (x, \pm \sqrt{1-x^2}, x \pm \sqrt{1 - x^2})^T = (\cos t, \sin t, \cos t + \sin t)^T \quad (***) $$
Looking at the extremal distances from the origin: $$ 0 = \frac{d}{dt}\lVert u(t) \rVert^2 = \frac{d}{dt} \left((\cos t)^2 + (\sin t)^2 + (\cos t + \sin t)^2\right) = \frac{d}{dt} 2 (\cos t \sin t) = 2(-(\sin t)^2 + (\cos t)^2) = 2(2(\cos t)^2 - 1) \iff \\ \cos t = \pm 1/\sqrt{2} \iff t \in \{ \pm \pi/4, \pm 3\pi/4 \} $$
The minor semi-axis is at $u_- = (1/\sqrt{2}, -1/\sqrt{2}, 0)^T$ with $\lVert u_- \rVert = 1$ and the major semi-axis at $u_+ = (1/\sqrt{2}, 1/\sqrt{2}, \sqrt{2})^T$ with $\lVert u_+ \rVert = \sqrt{3}$.
What is left is to ensure that $(***)$ is a parameterized ellipse. We would need to transform coordinates $u=(x,y,z)^T$ into local coordinates $v = (\xi, \eta, 0)^T$ within the plane.
We use $P (1/\sqrt{2}, -1/\sqrt{2}, 0)^T = (1, 0, 0)^T$ and $P(1/\sqrt{2}, 1/\sqrt{2}, \sqrt{2})^T = (0, \sqrt{3}, 0)^T$.
An orthogonal to the plane is $(1,0,1)^T \times (0,1,1)^T = (-1, -1, 1)^T$ which gives the normal $n = (-1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3})^T$. So we want $P(-1/\sqrt{3},-1/\sqrt{3}, 1/\sqrt{3})^T=(0,0,1)^T$. This gives $$ P = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} & -1/\sqrt{3} \\ -1/\sqrt{2} & 1/\sqrt{2} & -1/\sqrt{3} \\ 0 & \sqrt{2} & 1/\sqrt{3} \\ \end{matrix} \right)^{-1} $$
It turns out that $P u(t)$ describes a parameterized ellipse with semi-axes $1$ and $\sqrt{3}$.
Note: The images were created with GeoGebra and Gnuplot, some algebra was done and checked with Maxima.