For which values of $c$ are there solutions to the system of equations $x^2 + y^2 + z^2 = 1, x^2 + y^2 + c = z$?

Question

For which real values of $c$ are there real solutions to the system of equations $$x^2 + y^2 + z^2 = 1 \tag{1}$$ $$x^2 + y^2 + c = z \tag{2}$$

My work so far:

We can subtract the two equations and get $$z = \frac{-1 \pm \sqrt{5 +4c}}{2} \\ c \geq -\frac 5 4.$$

This is a necessary, but not sufficient, condition. Since $x^2 + y^2 \geq 0$, we can add the following additional conditions $$z^2 \ \leq 1 \\ z \geq c$$ and therefore $$c \leq 1.$$

Do all values of $c \in [-\frac 5 4, 1]$ have real solutions? For which values of $c$ are there two satisfying $z$ and for which only one? How do I prove these?

Answer 1

$c=z^2+z-1$ and the range of $z$ is exactly $[-1,1]$. Since $c$ as a function of $z$ is differentiable, $c$ obtains its extreme values either at the boundary points or where its derivative is $0$. Since it’s continuous, it obtains every point between its extremes. Because $[-1,1]$ is compact, it obtains its extrema on the interval. The derivative is $2z+1$ giving a potential extreme at $-1/2$. Testing $-1,-1/2,1$ gives $0,-5/4,1$, so the range of $c$ is exactly $[-5/4,1]$.

You can double check this without any calculus by noting that $c=(z+1/2)^2-5/4$, so this is always at least $-5/4$ and since $|z+1/2|\leq |z|+1/2\leq 3/2$, $c\leq 9/4-5/4=1$.

Answer 2

It would be easier to consider the problem geometrically. What you are asking is for what values of $c$ do the surfaces

$$x^2+y^2+z^2=1$$

the unit sphere centered at the origin, and

$$z = x^2+y^2+c$$

the circular paraboloid with vertex $(0,0,c)$, have any nonempty intersections?

Well, shifting the paraboloid vertex up is easy, clearly the last intersection exists at $c=1$, when only the vertex intersects the sphere. Then visualize the paraboloid sliding down from there. So long as the vertex is within the sphere, the intersection will always be a single circle.

Then, at $c=-1$, we get a circle and point again, and here as we keep sliding down we get a pair of circles. When do the circles vanish? It's when the paraboloid, in some loose sense, is "tangent" to the the sphere, and that's exactly when the pair of circles becomes one circle again. This corresponds to the doubled root you found in your quadratic equation for $z$.

Since you already did the algebraic work, let's organize the solutions geometrically as promised:

$\textbf{Case 1:}$ ($c=1$) The solution is a single point, $(0,0,1)$

$\textbf{Case 2:}$ ($-1<c<1$) The solution is a single circle,

$$\left\{(x,y,z)\Bbb{R}^3|z = \frac{\sqrt{5+4c}-1}{2}\:\cap\:x^2+y^2=\frac{\sqrt{5+4c}-1-2c}{2}\right\}$$

$\textbf{Case 3:}$ ($c=-1$) The solution is a single circle,

$$\left\{(x,y,z)\Bbb{R}^3|z = 0\:\cap\:x^2+y^2=1\right\}$$

and a single point, $(0,0,-1)$

$\textbf{Case 4:}$ ($-\frac{5}{4} < c < -1$) The solution is a pair of circles,

$$\left\{(x,y,z)\Bbb{R}^3|z = \frac{-1\pm\sqrt{5+4c}}{2}\:\cap\:x^2+y^2=\frac{-1\pm\sqrt{5+4c}-2c}{2}\right\}$$

$\textbf{Case 5:}$ ($c=-\frac{5}{4}$) The solution is a single circle,

$$\left\{(x,y,z)\Bbb{R}^3|z = -\frac{1}{2}\:\cap\:x^2+y^2=\frac{3}{4}\right\}$$

$\textbf{Case 6:}$ ($c\notin\left[-\frac{5}{4},1\right]$) There are no solutions because geometrically the shapes have exhausted their intersections by being too far away from each other.