Given a fixed circle "c" and a fixed point "A" (in the plane of the circle), draw the tangent to the circle at a variable point "X" (movable, but constrained to be on the circle), reflect "A" across the tangent to obtain point "A'", what is the equation of the geometric locus traced by "A'" as "X" performs a full rotation on the circle?
Take the circle given by $$(x,y)=(\cos t,\sin t)$$.
We can assume that the point $A$ is on the $X$ axis: $A(a,0)$.
For each $t\in[0,2\pi)$, the tangent line to the circle is $r_t:x\cos t+y\sin t=1$.
The perpendicular to $r_t$ that passes through $A$ is $$s_t:x\sin t-y\cos t=a\sin t$$
Now we compute $B=r_t\cap s_t$, I omit the details, since it is simply a 2x2 linear system: $$B(\cos t+a\sin^2 t,-a\sin t\cos t+\sin t)$$ Now, $$\overrightarrow {AB}=(\cos t+a\sin^2 t-a,\sin t-a\sin t\cos t)$$ Since $A'=A+2\overrightarrow{AB}$, $$A'(2\cos t+2a\sin^2t-a,2\sin t-2a\sin t\cos t)$$