Let $$P$$ and $$Q$$ be any points on the curves $$(x-1)^2 + (y+1)^2 = 1$$ and $$y = x^2$$, respectively. The distance between $$P$$ and $$Q$$ is minimum for some value of the abscissa of $$P$$ in the interval

Let $$Q(t,t^2)$$ be a point on the parabola $$y=x^2.$$

The circle $$(x-1)^2+(y+1)^2=1$$ has centre $$C(1,-1)$$ and radius $$1.$$

For minimum distance between the circle and parabola, we minimize the distance between $$C(1,-1)$$ and $$Q(t,t^2).$$

Let

$$D^2=(t-1)^2+(t^2+1)^2$$

$$=t^2-2t+1+t^4+2t^2+1$$

$$=t^4+3t^2-2t+2$$

Now differentiate:

$$\frac{d(D^2)}{dt}=4t^3+6t-2$$

For minimum distance,

$$4t^3+6t-2=0$$

$$2t^3+3t-1=0$$

Checking signs,

$$f\left(\frac14\right)=2\left(\frac1{64}\right)+\frac34-1<0$$

and

$$f\left(\frac12\right)=2\left(\frac18\right)+\frac32-1>0$$

Hence,

$$t\in\left(\frac14,\frac12\right)$$

Now point $$P$$ lies on the circle along the line joining $$C$$ and $$Q.$$

Hence,

$$P=C+\frac{Q-C}{|Q-C|}$$

Therefore, abscissa of $$P$$ is

$$x_P=1+\frac{t-1}{\sqrt{(t-1)^2+(t^2+1)^2}}$$

From

$$2t^3+3t-1=0,$$

$$1-t=2t(t^2+1)$$

Substituting,

$$x_P=1-\frac{2t(t^2+1)}{\sqrt{4t^2(t^2+1)^2+(t^2+1)^2}}$$

$$=1-\frac{2t}{\sqrt{4t^2+1}}$$

Now for

$$t\in\left(\frac14,\frac12\right),$$

$$x_P$$

lies between

$$1-\frac1{\sqrt5}$$

and

$$1-\frac1{\sqrt2}$$

Numerically,

$$0.29<x_P<0.55$$

Hence, the abscissa of $$P$$ lies in the interval

$$\boxed{\left(\frac14,\frac12\right)}$$.