Tangent to circle in 3D
A question in my analytical geometry book asked me to find a tangent line to a 3D circle which made me think about this. Can there be a meaning of "tangent line" in case the circle is in 3D or even a "tangent plane" ? Because if I consider definition of tangent to a point for a curve that it should only "touch" the curve at that point, then for a 3D circle, shouldn't it be "family of planes"?
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$\begingroup$The tangent at that point signifies the direction of the curve at that point and hence has to be in the same plane.
$\endgroup$ 1 $\begingroup$I would try to answer the question assuming it's for a sphere in 3D.
A sphere in 3D would be written as $(x-a)^2+(y-b)^2+(z-a)^2-r^2=0$ or simply the surface $f(x)=0\ where\ f(x):=(x-a)^2+(y-b)^2+(z-a)^2-r^2$. One would intuitively define the tangent to its surface at a point P by a set of lines which are tangent at P to a curve(lying in the surface) passing through P(note: line tangent to one curve might not be tangent to another curve).
Now let's write the curve as X(t) i.e. (as a parametric curve) and clearly it should satisfy $f(X(t))=0$, differentiating the given expression with respect to t by applying the chain rule, you would get $$\nabla f(X(t)).\frac{dX}{dt}(t)=0$$ suppose that the point where we wanted to find the tangent is at $t_0$,i.e., point= $X(t_0)$=P, so the above equation would give $$\nabla f(X(t_0)).\frac{dX}{dt}(t_0)=0$$which would mean that $\nabla f(X(t_0))$ is perpendicular to tangent of every curve along the surface at the point P(since X(t) was any arbitrary curve). This naturally leads us to suspect that all the tangents would then have to lie in a plane since they are are all perpendicular to the vector $\nabla f(X(t_0))$. To confirm that this is indeed a plane you can verify that for 2 orthogonal circles(tangent at the point of intersection are perpendicular) passing point P would give perpendicular tangents and hence their direction vectors would be linearly independent and would span this plane.
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