To draw the required circle, we first calculate its radius using the relation chord length = $2\sqrt{r^2 - d^2}$. Given chord length is 6 cm and perpendicular distance $d$ is 3 cm, we have $6 = 2\sqrt{r^2 - 3^2}$. Solving this, $3 = \sqrt{r^2 - 9}$, so $9 = r^2 - 9$, which gives $r^2 = 18$ and radius $r = 3\sqrt{2}$ cm.

Following the hint, we construct the circle as the circumcircle of a suitable triangle. A right-angled isosceles triangle with legs of 6 cm each is suitable. First, draw $\triangle ABC$ with $\angle B = 90^\circ$ and AB = BC = 6 cm. Next, find the midpoint O of the hypotenuse AC. Since the midpoint of the hypotenuse is the circumcentre for a right triangle, O is the centre. Draw the circle with centre O and radius OA ($3\sqrt{2}$ cm).

In this circle, AB is a chord of 6 cm. The perpendicular distance from the centre O to the chord AB is half the length of the other leg BC (as O is the midpoint of AC). Thus, the distance is $6 / 2 = 3$ cm, satisfying the given conditions.