Let $\mathbb{S}^3$ be the standard Euclidean sphere. Now we define shifted circle as $$\mathbb{S}^1_f = \left\{\left(\dfrac{x}{\sqrt{2}},\dfrac{y}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\right): x^2+y^2 = 1\right\}$$ which is a submanifold of $\mathbb{S}^3$. I want to find the equation of great circle that passes through $P\in \mathbb{S}^1_f$ and is perpendicular to the $\mathbb{S}^1_f$.
What I was trying to do
- Equation of the tangent space at the point $P$.
- Then finding the normal to the tangent space at the point $P$
- Now find out the equation of the plane passing through $P$ whose normal vector is given (from step 2)
- Intersect with the $\mathbb{S}^3$ and it will give the equation of the great circle.
But the above-described method was very lengthy to compute. So is there any other method to compute this in general? I know, in general, the equation of the great circle is determined by two orthogonal points $P$ and $Q$, but how to describe all which are perpendicular to $\mathbb{S}^1_f$?
Any help will be appreciated.
Thanks in advance!
Assume $x_{0}$ and $y_{0}$ are real numbers with $x_{0}^{2} + y_{0}^{2} = 1$, so that $v_{0} := \frac{1}{\sqrt{2}}(x_{0}, y_{0}, 1, 0)$ lies on the circle $S^{1}_{f}$. The vectors \begin{align*} v_{1} &= (-y_{0}, x_{0}, 0, 0), \\ v_{2} &= \tfrac{1}{\sqrt{2}}(x_{0}, y_{0}, -1, 0), \\ v_{3} &= (0, 0, 0, 1) \end{align*} are an orthonormal basis of $T_{v_{0}}S^{3}$ by inspection. The vector $v_{1}$ is tangent to the circle $S^{1}_{f}$, so $v_{2}$ and $v_{3}$ are perpendicular to the plane containing $S^{1}_{f}$. Your great circle may therefore be parametrized by $$ (x, y, z, w) = (\cos t)v_{2} + (\sin t)v_{3} = (\tfrac{1}{\sqrt{2}}x_{0}\cos t, \tfrac{1}{\sqrt{2}}y_{0}\cos t, -\tfrac{1}{\sqrt{2}}\cos t, \sin t). $$