prove that radius is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$ a circle is described to pass through the origin and to touch the lines $x=1,x+y=2$ then prove that the radius of the circle is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$
solution i try

length of perpendicular from $(1-r,2r-1)$ is $\displaystyle \frac{1-r+2r-1-2}{\sqrt{2}}=r^2$
$r=2(\sqrt{2}-1)$ which not satisfy $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$
How to find correct radious in that question

let $$x_M,y_M$$ the coordinates of the centre of the circle, and $r$ the searched radius, then we get
$$\left|\frac{x_M+y_M-2}{\sqrt{2}}\right|=r$$
$$|x_M-1|=r$$
$$x_M^2+y_M^2=r^2$$
can you finish?
