The plane $x+y+z=1$ intersects the curve $x^2+y^2+z^2=1$ at a circle $C$,find the radius and the center of this circle.

Question

The plane $x+y+z=1$ intersects the surface $x^2+y^2+z^2=1$ at a circle $C$,find the radius and the center of this circle.

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The points of the intersection of the two curves are the ones that satisfies the two equations,we know that: $$x^{2}+y^{2}+z^{2}=\left(x+y+z\right)^{2}-2\left(xy+xz+yz\right)$$

The points of the intersection are the points for which we have the following curve: $$xy+xz+yz=0$$ $$x\left(1-x\right)+\left(1-z^{2}-xz\right)=0$$

$$x^{2}+z^{2}+x+xz=1$$

But I have no idea how to find the circle,can someone help me?

Answer 1

The line $l=\{(t,t,t)\mid t\in\Bbb R\}$ is orthogonal to the plane $p=\{(x,y,z)\in\Bbb R^3\mid x+y+z=1\}$ and contains the origin, which is the center of the sphere. So, the center of your circle is the point at which $l$ and $p$ intersect, which is $C=\left(\frac13,\frac13,\frac13\right)$. And the radius is then the distance from $C$ to any point of the circle, such as $(1,0,0)$.

Answer 2

We have the plane $(u,v,1-u-v)$ and the intersection given by $$u^2+v^2+(1-u-v)^2=1.$$ The circle has equation $$u^2+uv+v^2=u+v$$ in the $uv$-plane. Substituting $u=\frac{p}{\sqrt{3}}+q,v=\frac{p}{\sqrt{3}}-q,$ we have $$p^2+q^2=\frac{2}{\sqrt{3}}p$$ or $$(\sqrt{3}p-1)^2+(\sqrt{3}q)^2=1,$$ thus let $p=\frac{s+1}{\sqrt{3}},q=\frac{t}{\sqrt{3}},$ or $$u=\frac{1+s}{3}+\frac{t}{\sqrt{3}}\\ v=\frac{1+s}{3}-\frac{t}{\sqrt{3}}.$$ The center of the circle is given by $(s,t)=(0,0)$ thus we obtain $(u,v)=\frac{1}{3}(1,1)$ or

$$(u,v,1-u-v)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$$