I'm trying to solve this task:
The luminous point is located at $(0, 0, 0)$. Determine the cone of the shadow cast by the ball: $(x-x_{0})^{2} + (y-y_{0})^{2} + (z-z_{0})^{2} \leq R^{2}$
if $x_{0}^{2} + y_{0}^{2} + y_{0}^{2} > R^{2}$
I tried applying the last condition to sphere equation and got:
$x^{2} + y^{2} + z^{2} < 2(xx_{0} + yy_{0} + zz_{0})$
I assume $x_{0}, y_{0}, z_{0}$ are parameters so partial derivatives with respect to them are:
$F'_{x_{0}} = -2x + 2x_{0} = 0$
$F'_{y_{0}} = -2y + 2y_{0} = 0$
$F'_{z_{0}} = -2z + 2z_{0} = 0$
I am not sure what to do next. Could somebody please give a hint?
Thanks in advance.
The task is apparently to find the cone with apex at the origin which is tangential to a sphere of radius $R$ centered at $(x_0,y_0,z_0)$. The cone axis is obviously the vector from $(0,0,0)$ through $(x_0,y_0,z_0)$ so only the half-angle $\phi$ of the cone is missing. If we draw a cross section through cone and circle where the axis is in the plane of the drawing, that angle is the angle inside the right triangle where $R$ is the length of the edge opposite to $\phi$, and the long edge (hypothenuse) is the distance from apex to circle center, $\sqrt{x_0^2+y_0^2+z_0^2}$. See e.g. https://www.researchgate.net/figure/In-a-circles-tangent-to-the-sector-from-it-interior-In-b-computing-total-curvature_fig4_220868972 or https://undergroundmathematics.org/thinking-about-geometry/r9694/solution . Obviously $\sin\phi=R/\sqrt{x_0^2+y_0^2+z_0^2}$.