I have a cone and a circle graphed in 3D (imagine that both graphs stop at the xy plane such that the base of the cone is at the xy plane). I want to make it such that the base of the cone has a diameter equal to the circle (basically such that the base of the cone is the circle while still maintaining the height of the cone, which will make the cone much more shallow). Can someone explain how I could do this?
The graph of the cone is $z=-\sqrt{(x-23.5)^2+y^2} + 22$ and the graph of the circle is $(x-23.5)^2 + y^2 = 111.6^2$
A general form for the cone (that has the central axis at $(x,y)=(23.5,0)$) is $$ (x-23.5)^2 + y^2 = (kz+c)^2 $$ for some constants $k,c$. Notice that the cone that you have currently defined has $k=1, c=-22$. The circle has $k=0, c=111.6$
Your wish is that at $z=0$, the cross-section of the cone coincides with the circle, so that the right-hand side is equal to $111.6^2$. That means $$ (k\cdot 0 + c) = 111.6^2 $$ We get $c= \pm 111.6$.
On the other hand, you want that the cone's top is at $z=22$. What corresponds to the top part of the cone? Answer: When the right-hand side is equal to zero. Therefore, this condition becomes $$ k \cdot 22 + c = 0 $$ Which value of $c$ do we plug in? Well, we know that we want to ultimately write the expression as $$ \sqrt{(x-23.5)^2 + y^2} = kz+c \geq 0 $$ and we want the expression to be valid in the range $0\leq z \leq 22$. Therefore, the sign of $c$ should be positive and $k$ should be negative.
Plugging in the solution, we get $$ (x-23.5)^2 + y^2 = \left(-\frac{116.6}{22} z + 111.6 \right)^2 $$ And if you want to "cut out" the part above $z=22$, you can take the square roots to obtain $$ \sqrt{(x-23.5)^2 + y^2} = -\frac{116.6}{22} z + 111.6 $$ (You can still re-arrange the terms if you want, but I believe this is a valid solution)