Deduce that the intersection of two graphs is a vertical circle

Question

Let C be the curve of intersection of the plane $x + y = 2b$ and the sphere $x^2 + y^2 + z^2 = 2b(x + y)$. Deduce that C represents a vertical circle centred on $(b,b,0)$ and having radius $√2b$.

My attempt:

Sketching the graphs and the intersection would make it obvious that the resulting curve is a vertical circle, however I am trying to explicitly get an equation of a circle.

First I rearranged the equation of the sphere to: $$(x-b)^2+(y-b)^2+z^2=2b^2$$ and I rewrote $x+y=2b$ to $(x-b)=-(y-b)$

Substituting in the equation of the sphere I get: $$(y-b)^2+(y-b)^2+z^2=2b^2$$ $$\implies 2(y-b)^2+z^2=2b^2$$

Which, I believe, is the projection of this vertical circle that I am trying to find on the yz-plane. But here I get stuck, and I ended up going in circles (no pun intended) if I attempted any more substitutions. I also tried to convert the equation of the sphere to parametric form but I still didn't manage.

How should I go about this? I should also add that I do not think this required a long answer due to being asked to "deduce" that the intersection is a circle and that it was an exam question.

Answer 1

Since you write about the radius being $\sqrt2\,b$, I will assume that $b>0$.

First of all, since the plan $x+y=2b$ is a vertical plane, whatever the intersection may be, it will be a vertical figure.

On the other hand if you intersect a plane with a sphere, then, assuming that the intersection has more than one point, what you get is always a circle.

In this specific case, the center of the sphere (which is $(b,b,0)$) belongs to plane. Therefore, the intersection between the sphere and the plane (which is $C$) will be a maximal circle of the sphere and it will have the same radius as well as the same center as the sphere. But the center of the sphere is located at $(b,b,0)$ and its radius is $\sqrt2\,b$.