Largest circle tangent at the hyperbola vertex

The inscribed circle in question shall have center $(0,1+r)$ and radius $r$ , which is represented as $x^2 + [y-(1+r)]^2 = r^2.$ If it intersects the hyperbola $y^2-x^2=1$ in just one point (namely $(x,y)=(0,1)$ ), then substituting $x^2=y^2-1$ into the circular equation yields the quadratic;

$(y^2-1) + [y^2 - 2(1+r)y + (1+r)^2] = r^2 \Rightarrow 2y^2 -2(1+r)y + 2r = 0$ ;

or $\large y = \frac{2(1+r) \pm \sqrt{4(1+r)^2 - 4(2)(2r)}}{4} = \frac{(1+r) \pm \sqrt{1+2r+r^2-4r}}{2} = \frac {(1+r) \pm \sqrt{(r-1)^2}}{2}$ .

There will be only one intersection point between the circle & the hyperbola iff the discriminant equals zero $\Rightarrow \boxed{r = 1}.$