(ii) $(0.2755,0.3803)$

(i) Take an arbitrary point $(x, y)$ on $C$. Distance between $(x, y)$ and $(1,0) = s =\sqrt{(x-1)^2+y^2} \Rightarrow s^2=(x-1)^2+y^2 $

(ii) Differentiating w.r.t. $x$ gives
$$2 s \frac{\mathrm{d} s}{\mathrm{~d} x}=2(x-1)+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}$$
When $s$ takes a minimum value, $\frac{\mathrm{d} s}{\mathrm{~d} x}=0$
Thus $$\begin{array}{l}
2(x-1)+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
x-1+\left(\frac{x}{1-x}\right)\left[\frac{1-x-x(-1)}{(1-x)^2}\right]=0 \\
\frac{x}{(1-x)^3}=1-x \\
x=(1-x)^4 \\
x^4-4 x^3+6 x^2-5 x+1=0 \\
\text { By GC (Poly Root Finder), } \\
x \approx 0.2755 \text { (to } 4 \text { d.p.) } \\
y=\frac{0.2755}{1-0.2755} \approx 0.3803
\end{array}$$
So the point on $C$ having a minimum distance from the point $(1,0)$ is $(0.2755,0.3803)$.

(iii)The minimum distance $=\sqrt{(0.2755-1)^2+0.3803^2} \approx 0.818$