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.

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.
