Finding the area of a circle that is formed by cutting a sphere. Say I have a sphere $x^2+y^2+z^2=a^2$ and a plane $x+y+z=b.$ How do I find the surface area of the circle cut by the sphere on the plane? I think I would use the surface integral and for graphs the surface "element" is $\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}$ where $f_x$ and $f_y$ are the derivatives of $x$ and $y$ accordingly. What would $f(x,y)$ be here exactly? Or should I project this circle onto $xOy$ plane, have $f(x,y)=b-x-y$. Either way I don't know how to project the surface onto the $xOy$ plane, where it is most likely a ellipse, where then I could parameterize the domain and go on from there.
I would really like to know how to project this onto $xOy$, because I can use this for other problems.
 A: By symmetry the center of the circle is $c=\left(\frac b3,\frac b3,\frac b3\right)$, as noted in the comments. By the Pythagoras's theorem the radius of the circle is
$$r=\sqrt{a^2-|c|^2}=\sqrt{a^2-\frac{1}{3}b^2}$$
whenever the expression is defined, otherwise there is no intersection between the sphere and the plane, and the area is zero. Therefore the area of the circle is
$$A_C=\pi r^2=\pi\left(a^2-\frac{1}{3}b^2\right).$$
You can obtain the same result projecting on the $xOy$ plane. Substitute for $z=b-x-y$ in the sphere equation to obtain
$$x^2+y^2+(b-x-y)^2=a^2. $$
Now we know that the projection will be an ellypse (every affine image of a circle is), with principal axes in direction parallel to $(1,1)$ and $(1,-1)$ (again by symmetry). Therefore we make the orthonormal change of variable $$\begin{cases}w=\frac{1}{\sqrt 2}(x+y)\\z=\frac{1}{\sqrt2}(x-y)\end{cases}$$
that makes the coordinate axes parallel to the principal axes of the ellypse, preserving areas, in order to obtain an equation of an ellypse in the normal form. The equation becomes
\begin{align}
&\frac12(w+z)^2+\frac12(w-z)^2+(b-\sqrt2 w)^2=a^2\\
& z^2+3w^2-2\sqrt2 bw+b^2=a^2\\
& z^2+3\left(w-\frac{\sqrt2}{3}b\right)^2=a^2-\frac13 b^2
\end{align}
which is the equation of an ellypse with center $(w,z)=\left(\frac{\sqrt 2}{3}b,0\right)$ and semiaxes $\ell_1=\frac{1}{\sqrt3}\sqrt{a^2-\frac13 b^2}$ and $\ell_2=\sqrt{a^2-\frac13 b^2}$, with area $A_E=\pi \ell_1\ell_2=\frac{\pi}{\sqrt 3}\left(a^2-\frac13 b^2\right)$, again under the condition that $a^2-\frac13 b^2$ be non negative.
Now the surface element of $f(w,z)=b-x-y=b-\sqrt 2 w$ is
$$\sqrt{1+f_w(w,z)^2+f_z(w,z)^2}=\sqrt{1+(-\sqrt2)^2}=\sqrt3$$
which integrated on the ellypse gives again
$$A_C=\sqrt3 A_E=\pi\left(a^2-\frac13 b^2\right).$$
