= Solution
Take a union of $d$ distinct lines in <general position for projective lines>, meaning that no three pass through one point. Each pair contributes one distinct <singular point>, and every other point lies on just one nonsingular line. Thus the union has <degree of a projective curve> $d$ and exactly
$$
\boxed{\binom d2=\frac{d(d-1)}2}
$$
distinct <singular points>. For $d=1$, this gives a line with zero <singular points>, as required.
For a concrete construction, choose distinct $t_1,\ldots,t_d\in\mathbb C$ and set
$$
L_i:\quad y=t_i x+t_i^2z,
\qquad
C:\quad\prod_{i=1}^d(y-t_ix-t_i^2z)=0.
$$
For $i\ne j$, the intersection is $[-(t_i+t_j):-t_it_j:1]$. Substitution into the equation of $L_k$ gives $-(t_k-t_i)(t_k-t_j)$, which is nonzero when $k$ is distinct from $i,j$. Thus there are no triple intersections, including at infinity.
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