Parametrize line given by intersection of sphere and plane

Question

Sphere of radius $1$ centred at $(x_0, y_0, z_0).$

The plane $z = cx$ where $c$ is a constant.

How do I find the parametrization of the intersection of the sphere and plane?

Answer 1

Without caring too much about the range of the parameters in the various equations, we can write a parametric equation for the sphere \begin{align} x &= x_0+\sin\theta\cos\phi,\\ y &= y_0+\sin\theta\sin\phi,\\ z &= z_0+\cos\theta \end{align} and a parametric equation for the plane \begin{align} x &= s,\\ y &= t,\\ z &= cs. \end{align} Comparing we have \begin{align} &x_0+\sin\theta\cos\phi = s,\\ &y_0+\sin\theta\sin\phi = t,\\ &z_0+\cos\theta = cs \end{align} from which we can express $s,t,\phi$ in terms of, for example, $\theta$: \begin{align} s &= \frac{1}{c}(z_0+\cos\theta),\\ \phi &= \arccos\left(\frac{s-x_0}{\sin\theta}\right) = \arccos\left(\frac{z_0+\cos\theta-cx_0}{c\sin\theta}\right),\\ t &= y_0+\sin\theta\sin\phi=y_0\pm\sin\theta\sqrt{1-\left(\frac{z_0+\cos\theta-cx_0}{c\sin\theta}\right)^2} \end{align} Finally, the parametric equation of the curve could be written as \begin{align} x &= \frac{1}{c}(z_0+\cos\theta),\\ y &= y_0\pm\sin\theta\sqrt{1-\left(\frac{z_0+\cos\theta-cx_0}{c\sin\theta}\right)^2},\\ z &= z_0+\cos\theta. \end{align}