Master Lock

We can always find an inscribed circle inside a rhombus, as shown above.

The diagonals of all rhombi intersect and bisect each other at one point, which is shown as point E in the picture, making AE = EC and BE = ED . Moreover, these diagonals are perpendicular to each other, so the four triangles ABE, BEC, CED, & DEA are all congruent and have the same areas, for a rhombus has equal sides (which act as bases of these triangles) and all 3 sides of the triangles are equal. With the same areas, that means these four triangles have the same height h, and when we draw these height lines passing to point E, point E will act as a center for an inscribed circle with a radius = h. In other words, the sides of the rhombus serve as the tangents of an inscribed circle.

As a result, we can always draw an inscribed circle within a rhombus.

However, a rhombus can not have a circumcircle because its two diagonals have different lengths, so when we try cycling it, the vertices of the shorter diagonal will never be able to touch the bigger circle, as shown below. Since all the rhombus vertices can't touch the circle, no circumcircles exist for any rhombus.

Opposite angles of a cyclic quadrilateral are supplementary. So rectangle and square or out. Diagonals do not bisect the angles so rectangle and ||gram all other are out except rhombus. Rhombus diagonals are angle bisectors, so will have an incircle, but opposite angles are not supplementary so no circumcircle.