I have been asked to show that the intersection of a sphere and a plane is a circle using two methods, at least this is my interpretation of a problem in a text.
The first method was something to do with hamilton's principle. It resulted in an integral from the pit of hell. I might ask a separate question about these integrals later.
What I am having trouble with at the moment is the more vanilla way of doing this.
I wrote down what I believe to be the equation of a plane.
$ax + by + cz + d = 0$
so in scalar form something familiar like:
$a(x - x_0) + b(y - y_0) + c(z-z_0) = 0$
I just then said the equation of the plane through (0,0,0) is
$x + y + z = 0$
I wrote the equation for a sphere similarly with the center at (0,0,0) to be
$x^2 + y^2 + z^2 = r^2$
I tried to solve for $z$ of the plane, and substitute it into the sphere equation in hopes of getting the equation of a circle.
Something that looks like
$(x-h)^2 + (y - k)^2 = r^2$
This did not happen.
I got
$2x^2 + 2xy + 2y^2 = r^2$
setting r to 1
$2(x^2 + xy +y^2) = 1$
It would seem I can rewrite this as
$(x + y)^2 + x^2 + y^2 = 1$
I don't think I have made any illegal moves. What's going on?
$ax+by+cz=0$ is the equation of a plane through the origin. The normal to that plane is $(a,b,c)$.
You've chosen a particular plane $x+y+z=0$ which is a plane through the origin with the normal pointing along the line defined by $x=y=z$. Then clearly $z=-x-y$ and you get $$x^2+y^2+(x+y)^2=r^2\tag{1}$$ as the equation of the circle, but the circle is in a plane at an angle of $45^\circ$ to any of the existing axes. To get that into the form $p^2+q^2=r^2$ you need to rotate $x'=x\cos45+y\sin45$, $y'=y\cos45-x\sin45$, then $z'=z\cos45+x'\sin45$, $x''=x'\cos45-z\sin45$, and then (1) will turn into an equation like $x''^2+y'^2=r^2$.
While that might be a useful exercise to do, you might have been better off picking a plane like $z=0$.