vectors
I know how to find a tangent plane to a sphere at any given point on the sphere.
But I have been given a tangent plane and a sphere, and need to prove, that the plane is indeed a tangent plane to a circle.
The circle has a radius of 5, center of $(-1,3,0)$ and equation of $(x+1)^2 + (y-3)^2 + z^2 = 25$.
The plane has an equation of $4x+3z+29=0$.
How to I prove that that the plane is a tangent plane first of all, and then how would i find the point at which it touches the circle?
Consider a cross-section of the figure. Then it takes one of the following two forms (here I have called one radius $R$, the other radius $r$, and the distance between the centers $D$)
Case 1: Plane does not pass the between the spheres.
In this case we can calculate $\cos\theta=\dfrac{R-r}{D}$, hence $\theta=\cos^{-1}\dfrac{R-r}{D}$
Case 2: Plane does pass between the spheres.
Similarly, here we can calculate $\sin\theta=\dfrac{R+r}{D}$, hence $\theta=\sin^{-1}\dfrac{R+r}{D}$. However, in this diagram we are not quite so interestd in $\theta$ as we are in $\frac{\pi}{2}-\theta$, the angle of the vector from the center of the first circle perpendicular to the plane. Calling this angle $y$, we have $y=\cos^{-1}\dfrac{R+r}{D}$.
Now we turn to our question. For convenience we will let the center of the first sphere be $(0,0,0)$ and the center of the second sphere as $(D,0,0)$ in $\textit{spherical}$ coordinates ($(x,\theta,\phi)$ meaning distance $x$ from the origin, angle $\theta$ from the $z$-axis, and angle $\phi$ on the projection to the xy-plane (see the first picture here)).
Then for case 1, we can describe a desired plane as the normal plane to the vector $(r,\cos^{-1}\dfrac{R-r}{D},\phi)$ and passing through the point $(r,\cos^{-1}\dfrac{R-r}{D},\phi)$ for $\phi\in[0,2\pi)$.
Similarly for case 2, we can describe a desired plane as the normal plane to the vector $(r,\cos^{-1}\dfrac{R+r}{D},\phi)$ and passing through the point $(r,\cos^{-1}\dfrac{R+r}{D},\phi)$ for $\phi\in[0,2\pi)$.
These are exactly all planes tangent two the two spheres.
Please note for the entirety of this we have assumed $D>R+r$ (or in other words no sphere contains part of the other).
The plane unit normal vector is $(1,2,2)$ normalized or $n=(1/3,2/3,2/3)$. Draw a line $l$ through the origin in the direction of $n.$ This line intersects the unit sphere at the point closest to the plane: $(1/3,2/3,2/3).$ (The line $l$ also intersects the unit sphere at $(-1/3,-2/3,-2/3),$ but this is the point on the sphere farthest from the plane)
The reason this is the closest point on the sphere to the plane is that the line $l$ is orthogonal to the tangent plane of the sphere at the point where it intersects the sphere and also orthogonal to the plane.
Best Answer
Let $f(x,y,z) = (x+1)^2 + (y-3)^2+z^2$. The gradient of $f$ is normal to the surface of the sphere. Meanwhile the normal to the plane is $(4,0,3)$. So find $x$, $y$, and $z$ values for which $\nabla f$ is a multiple of $(4, 0, 3)$. Then set the multiple so that that these $x$, $y$, and $z$ values lie on the surface of the sphere.