<s>[INST] Newbie: Find the intersection of a line and a circle and interpret geometrically Find the points of intersection of the line $x+y+k=0$ and the circle $x^2+y^2=2x$. Show that there are two points of intersection if: $-1-\sqrt2<k<-1+\sqrt2$, one point of intersection if: $k=-1\pm\sqrt2$, and none otherwise. Interpret the results geometrically. What is the shortest distance between the point (1,0) and either of the lines $x+y-1\pm\sqrt2=0$
My attempt:
$x + y=0$ 
$y=-x-k$  
$x^2+y^2=2x$
$x^2+(-x-k)^2=2x$
$x^2+(x^2+2xk=k^2)=2x$
$2x^2 -2x+2xk+k^2=0$
$2x^2+(-2+2k)x +k^2=0$ 
Following what I read here: How do I calculate the intersection(s) of a straight line and a circle? I substituted into the quadratic equation and made the assumption that if $b^2 -4ac=0$ then there would be a single point of intersection.
$0=b^2-4ac$
$0=(-2+2k)^2-4(2)(k^2)$
$0=-4k^2-8k+4$
$0=k^2+2k-1$
$0=(k-1)^2$
The parameters for k, as stated in the question, are clearly not met so I'm not sure what else to try.
 [/INST] Everything is correct up until your last two lines. You seem to be assuming that $k^2 + 2k -1$ is equal to $(k-1)^2$, which is not true.
Your equation $k^2 + 2k -1 = 0$ is correct, and it has roots $k = -1 \pm \sqrt2$, which is probably what you were expecting.
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