Choosing a value so a line and circle intersect a one, two, and no points Let l be a line and C be a circle.
$y=x+d$, where $d$ is to be determined.
$C=x^2+y^2=4$


*

*Pick a value for $d$ so that l and C intersect at one point.  

*Pick a value for $d$ so that l and C intersect at two distinct points. 

*Pick a value for $d$ so that l and C do not intersect.


Am I able to use this: How do I calculate the intersection(s) of a straight line and a circle??
Graphically, I could answer B and C.  I'm hoping there's an algebraic way to solve all three parts.

For intersection points both equations
$$
y = x + d \\
x^2 + y^2 = 2^2
$$
must hold. Inserting the first into the second we get
$$
y = x + d
$$
and
$$
2^2 = x^2 + (x+ d)^2 = 2 x^2 + 2xd + d^2 \iff \\
2 = x^2 + x d + d^2/2 = (x + d/2)^2 - d^2/4 + d^2/2 \iff \\
2 - d^2/4 = (x + d/2)^2 \iff \\
x = \frac{-d \pm \sqrt{8 - d^2}}{2}
$$
Which gives the solutions
$$
P = 
\left( 
\frac{-d \pm \sqrt{8 - d^2}}{2},
\frac{-d \pm \sqrt{8 - d^2}}{2} + d 
\right)
$$
The number of real solutions is determined by the argument of the square root
$$
\Delta(d) = 8 - d^2
$$
If $\Delta(d) = 0$ or $d = \pm\sqrt{8} = \pm 2\sqrt{2}$ there is one solution.
If $\Delta(d) > 0$ or $\lvert d \rvert < 2\sqrt{2}$ there are two solutions.
If $\Delta(d) < 0$ or $\lvert d \rvert > 2\sqrt{2}$ there is no (real) solution.

You can fiddle with it here.
