<s>[INST] 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.
 [/INST] Information

We have two equations:


*

*$ y = x + d $

*$ x^2 + y^2 = 4 $
We must find the number of intersections depending on $d$.

Solution

Step 1
Let us begin with finding the relation between $x$ and $d$.
Why? We need to see how we need to manipulate $d$ to get 0, 1, or 2 valid $x$ values. Each valid $x$ value will (obviously) correspond to a point of intersection.
$ x^2 + y^2 = 4 $
$ x^2 + (x + d)^2 = 4 $
$ (2)x^2 + (2d)x + (d^2 - 4) = 0 $
$ x = \frac{-(2d) \pm \sqrt{(2d)^2 - 4(2)(d^2 - 4)}}{2(2)} 
= \frac{-d \pm \sqrt{8 - d^2}}{2} $

Step 2
Now we shall analyze our relation between $x$ and $d$.
Let 
$ x = \frac{-d \pm \sqrt{A}}{2} $
where 
$ A = 8 - d^2 $.

Step 2A
If $ A = 0 $ then $x$ has one solution.
What must $d$ be such that $A = 0$?
$ A = 8 - d^2 = 0 $
$ d^2 = 8 $
$ d = \pm 2\sqrt{2} $

Step 2B
If $ A > 0 $ then $x$ has 2 solutions.
What must $d$ be such that $A > 0$?
$ A = 8 - d^2 > 0 $
$ d^2 < 8 $
$ d < \pm 2\sqrt{2} $
$ -2\sqrt{2} < d < 2\sqrt{2} $

Step 2C
If $ A < 0 $ then $x$ has 0 solutions.
What must $d$ be such that $A < 0$?
$ A = 8 - d^2 < 0 $
$ d^2 > 8 $
$ d > \pm 2\sqrt{2} $
$ d < -2\sqrt{2} $ or $ 2\sqrt{2} < d $

Answer


*

*0 solutions: $ d < -2\sqrt{2} $ or $ 2\sqrt{2} < d $

*1 solution: $ d = \pm 2\sqrt{2} $

*2 solutions: $ -2\sqrt{2} < d < 2\sqrt{2} $

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