## Task A-4.4.

A circle passes through the points $A(0,5)$ and $B(0,-1)$, and its center lies on the line $y=2x-6$. Determine the sine of the inscribed angle over the smaller arc $\overparen{A B}$ of this circle.
## Second Solution.

Notice that the point $M(0,2)$ is the midpoint of the segment $\overline{A B}$, and that this segment lies on the $y$-axis. The center of this circle is located on the perpendicular bisector of the segment $\overline{A B}$. This is a line that is perpendicular to the $y$-axis and passes through the midpoint, so its equation is $y=2$.

By substituting the condition $y=2$ into the equation of the line $y=2 x-6$, we obtain the coordinates of the center of the circle $O(4,2)$.

The radius of the circle is $|A O|=\sqrt{(0-4)^{2}+(5-2)^{2}}=5$.

The sine of the inscribed angle over the smaller and the larger arc $\overparen{A B}$ of this circle are the same (since $\sin x=\sin (\pi-x))$. The inscribed angle over the larger arc is equal to half of the central angle $\varangle A O B$ over the chord $\overline{A B}$. Since $M$ is the midpoint of the base of the isosceles triangle $A B O$, we conclude that the sine of the desired angle can be calculated as the sine of the angle $\varangle A O M$.

Triangle AMO is a right triangle, so we have

$$
\sin \varangle A O M=\frac{|A M|}{|A O|}=\frac{\sqrt{(0-0)^{2}+(5-2)^{2}}}{5}=\frac{3}{5}
$$

Note: Elements of the above solutions can be combined. Generally, in all solutions, the structure of scoring is as follows: finding the center of the circle is worth 3 points, calculating the radius of the circle is worth 1 point, determining the method of calculating the desired value of the problem is worth 1 point, and the final calculation is worth 1 point.