Problem 1661

1. Define the Problem and Setup:
Let $\omega_1$ and $\omega_2$ be two orthogonal circles with centers $O_1$ and $O_2$ respectively. Let $O$ be the center of $\omega_1$. Consider a diameter $AB$ of $\omega_1$ such that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at points $F$ and $G$. We need to prove that $FGOB$ is cyclic.

2. Introduce Inversion:
To simplify the problem, we use inversion with respect to $\omega_1$. Since $\omega_2$ is orthogonal to $\omega_1$, $\omega_2$ remains unchanged under this inversion. Points $A$ and $B$ lie on $\omega_1$, so they are preserved under inversion.

3. Analyze the Inversion:
Under inversion, the circles $\Gamma_1$ and $\Gamma_2$ (the circumcircles of $\triangle AOF$ and $\triangle AOG$ respectively) are mapped to lines $AF_2$ and $AG_2$ which are tangents to $\omega_2$. Here, $F_2$ and $G_2$ are the images of $F$ and $G$ under inversion.

4. Properties of Inversion:
Since $O$ is the center of inversion, it maps to the point at infinity. Therefore, we need to show that $F_2BG_2$ is a straight line. This would imply that $F_2$, $B$, and $G_2$ are collinear.

5. Use of Polar Concept:
Note that $F_2G_2$ is the polar of $A$ with respect to $\omega_2$. To prove that $B$ lies on the polar of $A$ with respect to $\omega_2$, we need to show that $B$ lies on the line $F_2G_2$.

6. Intersection Points and Power of a Point:
Let $AB$ intersect $\omega_2$ at points $M$ and $N$, where $M$ is closer to $A$ than $N$. Since $\omega_2$ is orthogonal to $\omega_1$, the power of point $O$ with respect to $\omega_2$ is given by:
$$OB^2 = OM \times ON$$
This relationship confirms that $B$ lies on the polar of $A$ with respect to $\omega_2$.

7. Conclusion:
Since $B$ lies on the polar of $A$ with respect to $\omega_2$, and $F_2G_2$ is the polar of $A$, it follows that $F_2$, $B$, and $G_2$ are collinear. Therefore, $FGOB$ is cyclic.

$$\boxed{\text{None}}$$