<s>[INST] find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one.  I'm teaching my self pre-calc out of a book and it isn't showing me how to do this.  I've been all over the internet and could only find a few examples.  I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in.  What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values.  These points are not on the line.  Can someone tell me what I'm doing wrong please?
 [/INST] Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $\frac32\pm\frac{\sqrt 7}2$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(\frac32\pm\frac{\sqrt 7}2,-\frac12\pm\frac{\sqrt 7}2)$
</s>