To address the two parts of the problem step by step:

1. Prove that triangle $\triangle MON$ is right at $O$:

• Since $AB$ is a diameter of the circle and $O$ is the center, the circle's radius is $OA = OB$.
• The tangent at $P$ to the circle intersects the tangents at $A$ and $B$ at $M$ and $N$, respectively.
• Tangents drawn from an external point are perpendicular to the radius drawn to the point of tangency.
  • $\angle OMA = 90^\circ$ and $\angle ONB = 90^\circ$, since $OA$ and $OB$ are radii perpendicular to their respective tangents.
• $OM$ and $ON$ are line segments connecting $O$ to points $M$ and $N$.
• Therefore, the triangle $\triangle MON$ contains $\angle MON = 90^\circ$, proving it is a right triangle at $O$.

2. Show that the circle circumscribed about $\triangle MON$ is tangent to the line $AB$:

• The circumscribed circle of $\triangle MON$ has its center at the midpoint of the hypotenuse $MN$, and its radius is half the length of $MN$.
• Since $\angle MON = 90^\circ$, the hypotenuse $MN$ is a diameter of the circumscribed circle of $\triangle MON$.
• $M$ and $N$ lie on tangents to the circle at $A$ and $B$, so the line $AB$ is parallel to the hypotenuse $MN$.
• The distance from $O$ (the center of the original circle) to the line $AB$ equals the radius of the circumscribed circle of $\triangle MON$.
• Therefore, the circle circumscribed about $\triangle MON$ is tangent to the line $AB$.

Do you need further elaboration or diagrams to delve deeper? Here are some related questions:

1. Why does $\angle MON$ being $90^\circ$ imply a unique circumscribed circle for $\triangle MON$?
2. What is the relationship between the tangents and the diameter $AB$?
3. Can the circumcircle's radius be expressed in terms of the radius of the original circle?
4. How does the parallel relationship between $AB$ and $MN$ influence tangency?
5. How would the proof change if $AB$ were not the diameter?

Tip: Drawing a clear diagram and marking all tangents and radii simplifies the proof significantly.