### Instruction: For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?### Response: I want to find out when the circle $x^2+y^2=k^2$ and the hyperbola $xy = k$ do not have any common points.
One way to approach this problem is to try to solve the system of equations by eliminating one of the variables and getting a quadratic equation in the other.
For example, if I solve for $x$ from the hyperbola equation, I get $x = \frac{k}{y}$.
Then I can substitute this into the circle equation and get $\frac{k^2}{y^2} + y^2 = k^2$.
Multiplying both sides by $y^2$, I get $k^2 + y^4 = k^2 y^2$.
Rearranging, I get $y^4 - k^2 y^2 + k^2 = 0$.
This is a quadratic equation in $y^2$, so I can use the quadratic formula to find its roots.
The roots are $y^2 = \frac{k^2 \pm \sqrt{k^4 - 4k^2}}{2}$.
In order for $y^2$ to be a real number, the discriminant $k^4 - 4k^2$ must be non-negative.
This means that $k^4 - 4k^2 \geq 0$.
Factoring, I get $k^2 (k^2 - 4) \geq 0$.
This implies that $k^2 \geq 4$ or $k^2 \leq 0$.
The latter case is impossible, since $k^2$ is always non-negative.
So the only condition is $k^2 \geq 4$.
This means that $k$ can be any integer whose absolute value is at least 2.
The integers that do not satisfy this condition are $k = -1, 0, 1$.
These are the only values of $k$ for which the graphs do not intersect, since for any other value of $k$, there will be at least one real solution for $y^2$.
Therefore, the answer is 3.