Step 1: Let $P(a\cos\theta+b\sin\theta,\ 0)$ and $Q(0,\ a\sin\theta-b\cos\theta)$.
Step 2: Distance formula: $$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Step 3: $PQ=\sqrt{(0-(a\cos\theta+b\sin\theta))^2+((a\sin\theta-b\cos\theta)-0)^2}$
Step 4: Expand both squares:
$$(a\cos\theta+b\sin\theta)^2 = a^2\cos^2\theta + 2ab\sin\theta\cos\theta + b^2\sin^2\theta$$
$$(a\sin\theta-b\cos\theta)^2 = a^2\sin^2\theta - 2ab\sin\theta\cos\theta + b^2\cos^2\theta$$
Step 5: Adding, the middle terms cancel:
$$a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)=a^2+b^2$$
Step 6: Hence $PQ=\sqrt{a^2+b^2}$.
Answer: (A)