The circle centre can be inside, on, or outside of the inscribed cyclic quadrilateral.
In the attached sketch,
for the cyclic quadrilateral abcd, Q+S=P+R=180 degrees.
If "d" is outside the circle on the line ad, angle S decreases, while Q remains the same. If "d" is inside the circle on the same line, S increases while Q remains the same.
In both these cases, Q+S is no longer 180 degrees.
Neither is P+R.
All angles sum to 360 degrees, but opposite angles will no longer sum to 180 degrees. Hence, if opposite angles sum to 180 degrees, the quadrilateral is cyclic, as a circle circumference will pass through all 4 vertices.
Wow! I really started off on the wrong foot. I have good reason for being a bit confused and tired [medicines], but have not made mistakes like this for a while. Sorry about that. It won't happen again. I was somehow thinking "rectangle" instead of "quadrilateral". In future I'll stick to answering when not so tired out.
Let me try again:
It does not matter if the quadrilateral includes the center or not. The proof is the same. With cyclic quadrilateral ABCD, the points A and C (or "arc AC") subtend angles at B and D. Each of those is one half of the angle subtended at the center [whether inside or out does not matter]. The angles at the center add to 360, so those at the circumference, B and D, add to 90. So then do A and C for several reasons, one of which is that the angles in any quadrilateral add to 360 in total.
EDIT: Just learning to upload an attachment. Please bear with me. This, if it works, is what I'm driving at: