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:

$x + y=0$


$2x^2 -2x+2xk+k^2=0$
$2x^2+(-2+2k)x +k^2=0$

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.


The parameters for k, as stated in the question, are clearly not met so I'm not sure what else to try.

  • $\begingroup$ I tried the following: $\endgroup$ – Lucian Feb 22 '14 at 16:23
  • $\begingroup$ You are doing the right thing the only problem is that you messed up the calculation, everything is correct up until: $-4k^2 - 8k +4=0$. It should be $4+4k^2-8k-8k^2=0$. $\endgroup$ – Ben Feb 22 '14 at 18:34
  • $\begingroup$ Yes I accidentally neglected to include that step in my above workings however if $4+4k^2-8k-8k^2=0$ is simplified, it can be expressed as $-4k^2-8k+4=0$ unless I'm mistaken. $\endgroup$ – Lucian Feb 22 '14 at 19:23
  • $\begingroup$ How do you find that there are two points of intersection if $-1-\sqrt2<k<-1+\sqrt2\,$ ? $\endgroup$ – Tony Piccolo Feb 23 '14 at 7:51
  • $\begingroup$ The link included in the question indicates that: "if $b^2+4ac>0$ then the line meets the circle in two distinct points." I presume that when $k=-1\pm\sqrt2$ that the line is tangential to the circle and if $k$ lies between these two values then it is crossing the circle at two points. $\endgroup$ – Lucian Feb 23 '14 at 9:12

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.

  • $\begingroup$ Oh yes now I see where I went wrong slams hand into face. Thanks for pointing out my idiotic factorising mistake. $\endgroup$ – Lucian Feb 23 '14 at 9:05

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