# Intersection of a sphere and plane problem

Hello,

I was wondering if anyone could offer some advice on this one. I have a multi-part problem, in which I can't get the second part. It starts like this:

(a)
Find the equation of the sphere passes through the point (6,-2,3) and has a center of (-1,2,1).

So I did this find and came up with:

(x+1)²+(y-2)²+(z-1)² = 69

(b)
Find the curve in which this sphere intersects the yz-plane.

So this is where I am stuck...

quasar987
Homework Helper
Gold Member
The zy plane is the set $\Pi_{zy}$ of all points (x,y,z ) of R^3 for which x=0.

and your sphere (I havent checked if you found the correct radius) is the set $\mathbb{S}_{\sqrt{69}}$ of all points (x,y,z) of R^3 that obey the equation (x+1)²+(y-2)²+(z-1)² = 69.

So, if you define the set $\mbox{Intersection}(\Pi_{zy},\mathbb{S}_{\sqrt{69}})$ of as the points (x,y,z) of R^3 that obey both to the x=0 and the (x+1)²+(y-2)²+(z-1)² = 69 condition, then points of this set are both the plane and the sphere. Logical?

Last edited:
Hmm.. This is a little confusing for me. Sorry about that. So... I am not sure what the notation $\Pi_{zy}$ means, but I am guessing that it means the set of points on a sphere in the zy plane? But what you are basically telling me to do, or to think about, is to let x = 0, then I can solve a system of 2 equations and two unknowns to solve for y and z when x = 0. How does that sound?

HallsofIvy
(y- 2)2+ (z- 1)2= 68. That's the equation for a circle. A standard parametrization for a circle is to let $\theta$ be the angle a radius makes with an axis. In particular, what must y and z be, in terms of $\theta$ so that your equation becomes
$68cos^2(\theta)+ 68sin^2(\theta)= 68$?