What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$ What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$

The lines $x+y=2$ and $x-y=2$ are perpendicular to each other and the circle is touching both the lines,these lines are tangents to the circle.Let points of tangency be $P$ and $Q$,let the center of the circle be $O$ and let the point where the lines $x\pm y=2$ meet be $R$.$OP$ is perpendicular to $PR$ and $OQ$ is perpendicular to $QR$,therefore $OPRQ$ is a square and the point $(-1,1)$ does not lie on any of the lines $x\pm y=2$.
But now i am stuck,how to solve further.Please help me. 

Let $(h,k)$ be the center of the required circle. 
The perpendicular distance from $(h,k)$ to $x+y-2=0$
=The perpendicular distance from $(h,k)$ to $x-y-2=0$
=The distance between $(h,k)$ to $(-1,1)$. 
The first two equality gives $|\frac{h+k-2}{\sqrt 2}|=|\frac{h-k-2}{\sqrt 2}|$, which will imply either $k=0$ or $h=2$
So, you have two cases, 
Case,I $k=0$
From the seond and third equality gives, 
$\frac{|h-2|}{\sqrt 2}=\sqrt{(h+1)^2+(1)^2}$
squaring on both sides, 
or $ \frac{(h-2)^2}{2}=(h+1)^2+1$
or $h^2-4h+4=2h^2+4h+2+2$
or $h^2+8h=0$
0r $h=0,-8$
Radius$=\sqrt{(-8+1)^2+(1)^2}=5\sqrt 2 or \sqrt{(0+1)^2+(1)^2}=\sqrt 2$   
Case-II, $h=2$
$\frac{|2+k-2|}{\sqrt 2}=\sqrt{(2+1)^2+(k-1)^2}$
squaring on both sides, 
or, $k^2= 18+2k^2-4k+2$
or,  $k^2-4k+30$
This doesn't have real solutions. So case-II is not valid.  
