4. If $N=4 a^{2}-12 a b+13 b^{2}-6 a+4 b+13$, then the value of $N$ is ( ).
(A) Positive
(B) Zero
(C) Non-negative
(D) Negative
Answer: (A)
Solution: The original expression $=\left(a^{2}-6 a+9\right)+\left(b^{2}+4 b+4\right)+3\left(a^{2}-4 a b+4 b^{2}\right)$
$$
=(a-3)^{2}+(b+2)^{2}+3(a-2 b)^{2} \geq 0 \text {, }
$$

Since $a-3, b+2, a-2 b$ cannot all be 0 at the same time,

Therefore, $\mathrm{N}$ is a positive number.
Hence, the answer is (A).