Unit Vector of a Sphere Not Centered at Origin in Cartesian Coordinates

Question

I have the equation $(x-2)^2+(y-1)^2+z^2=25$, and I want to find a normal to this surface (pointing outwards), at the point $(-2,4,0)$.

I tried using the gradient, but that seems to give me a vector that would work for a sphere centered at the origin. I know that for a sphere centered at the origin, the normal vector at a point on the surface is just the coordinates of that point, but the centering of the sphere is what is throwing me off here.

Answer 1

Gradients can obtain normals for not only arbitrarily centred spheres, but arbitrary surfaces for which the gradient is defined. Since $f(x,\,y,\,z)=c$ implies $\nabla f\cdot dx=df=0$, the normal to $f=c$ is in direction of $\nabla f$, thereby orthogonal to each $dx$ in the surface. In this case, the normal is proportional to $-8\vec{i}+6\vec{j}$, so the choices of unit normal are $\mp\tfrac45\vec{i}\pm\tfrac35\vec{j}$.

Answer 2

You can answer this question by thinking about the geometry; no need for gradients.

Translate the sphere back to the origin, along with the point on it you care about. The normal vector there will be the same as the normal vector to the original sphere at the original point.

Answer 3

The center of the sphere is $(2, 1, 0)$.

The normal is the vector that joins $(-2,4,0)$ to the center, that is $(-4,3,0)$.

If you want the unit normal, then it is $\left(-\frac{4}{5},\frac{3}{5},0\right)$.

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