1. Interpret the curves

   • Point $P$ lies on the circle $(x-1)^2+(y+1)^2=1$ which has center $C=(1,-1)$ and radius $1$.

   • Point $Q$ lies on the parabola $y=x^2.$

   We need the point $P$ on the circle for which the distance to the parabola is minimum.

2. Geometric idea for minimum distance between two smooth curves

   If the minimum distance occurs at points $P$ and $Q$ on the two curves, then the segment $PQ$ is along the common normal to both curves.

   Also, for a circle, the normal at $P$ is along the radius $CP$. Hence at the minimizing position, the line joining $P$ to $Q$ must pass through the center $C=(1,-1)$.

   Therefore, if $Q=(t,t^2)$ is the corresponding point on the parabola, then the normal to the parabola at $Q$ must pass through $(1,-1)$.

3. Equation of normal to the parabola $y=x^2$

   Let $Q=(t,t^2)$ on $y=x^2$.

   Slope of tangent at $Q$ is $\frac{dy}{dx}=2t.$

   So slope of normal is $-\frac{1}{2t} \quad (t\neq 0).$

   Equation of the normal at $Q$ is $y-t^2=-\frac{1}{2t}(x-t).$

   Since this normal passes through $(1,-1)$, substitute $x=1$, $y=-1$: $-1-t^2=-\frac{1}{2t}(1-t).$

   Multiply by $2t$: $-2t-2t^3=-(1-t).$

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

4. Solve the cubic

   We test $t=\frac12$:

   =\frac14+\frac32-1 =\frac34\neq 0. $$ Test $t=\frac13$: $$ 2\left(\frac{1}{27}\right)+3\left(\frac13\right)-1 =\frac{2}{27}+1-1 =\frac{2}{27}\neq 0. $$ Test $t\approx 0.3$: $$ 2(0.027)+0.9-1=-0.046. $$ Test $t\approx 0.32$: $$ 2(0.032768)+0.96-1=0.025536. $$ So the root lies near $t\approx 0.31$. We now find the corresponding point $P$ on the circle.

5. Find $P$ from the center toward $Q$

   Since $P$ is on the circle and lies on the line from center $C=(1,-1)$ toward $Q=(t,t^2)$, we have $P=C+\frac{Q-C}{|Q-C|}.$

   Here $Q-C=(t-1,t^2+1).$

   The abscissa of $P$ is $x_P=1+\frac{t-1}{\sqrt{(t-1)^2+(t^2+1)^2}}.$

   Using $t\approx 0.31$: $t-1\approx -0.69, \qquad t^2+1\approx 1.0961.$

   Then

   \approx \sqrt{0.4761+1.2014} \approx \sqrt{1.6775} \approx 1.295. $$ Therefore, $$ x_P\approx 1-\frac{0.69}{1.295} \approx 1-0.533 \approx 0.467. $$ Thus the abscissa of $P$ lies in $$ \left(\frac14,\frac12\right). $$

6. Check options

   • A: $\left(0,\frac14\right)$ ❌
   • B: $\left(\frac12,\frac34\right)$ ❌
   • C: $\left(\frac14,\frac12\right)$ ✅
   • D: $\left(\frac34,1\right)$ ❌

So the correct option is C.

7. Comparison with stored answer

   Stored correct answer: C

   Our derived answer: C

   Hence, they agree.