circle of inversion

Determine the equation of the circle reflection of the line $x = 2$ if the circle of reflection is $x^2 + y^2 + 2x = 0$ which in standard form is $(x+1)^2+y^2=1$ where $radius=1$ and center is $(-1,0)$.

Since $x=2$ is on the outside of the circle of reflection I am not sure what to do.

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The vertical line $x=2$ does not intersect the circle with radius $1$ and center at $(-1,0)$. – Sigur May 6 '13 at 0:12
I don't understand if you want to invert the line $x=2$ using geometric inversion by the circle or if you want to reflect the circle on the line. – Sigur May 6 '13 at 0:14
I believe invert the line $x=2$. Yeah I completely screwed up. Since $x=2$ is on the outside of the circle the inverse image should be inside $x^2+2x+y^2=0$ – user60887 May 6 '13 at 0:16

Since the circle of inversion has radius $1$ and is centered at $(-1,0)$ the line $x=2$ does not intersect the circle. Furthermore, the point $(2,0)$ is inverted to a point inside the circle.
Since the inversion of the line is a circle containing the center of inversion (i.e. $(-1,0)$) you only need to invert the point $(2,0)$ to obtain the equation of the circle coming from the inversion of the line.
So basically find the equation of the circle going through the point of inversion of $(2,0)$ and $(-1,0)$? – user60887 May 6 '13 at 0:27
@user60887, yes, that's it. You could choose any other point on the line to invert but that one is the easiest one since its inversion would also be on the $x$-axis. – Sigur May 6 '13 at 0:29
No. You have to use the center of inversion, that is, $(-1,0)$, instead of the origin $O=(0,0)$. – Sigur May 6 '13 at 0:38
If the center of inversion (that is, the center of the circle used to inversion) is $Q$ and the radius of this circle is $r$ than the inversion of a point $B$ is the point $B'$ which satisfies $d(B,Q)d(B',Q)=r^2$. – Sigur May 6 '13 at 0:41
So I get $(-2/3,0)$. What I did was just the distance formula. $1/3=$\sqrt[2]{(x_2+1)^2+(y_2)^2}. Since we know $y_2=0$ we just solve for $x_2$ which I got $(1/9)=(x_2+1)^2$. Solving for x I got two solutions but since the distance is positive I get $(-2/3,0)$. – user60887 May 6 '13 at 0:59