## Task B-4.2.

Determine the equation of the circle that is tangent to the line $x-3y+4=0$, and its center lies on the lines $2x-3y-9=0$ and $y+1=0$.
## Solution.

The center of the circle is at the intersection of the lines $2 x-3 y-9=0$ and $y+1=0$, which is the solution to the given system, and that point is $(3,-1)$.

(1 point)

The radius is equal to the distance from the center $S(3,-1)$ to the line $x-3 y+4=0$

$$
r=\frac{|3+3+4|}{\sqrt{1+9}}=\sqrt{10}
$$

Solution: $(x-3)^{2}+(y+1)^{2}=10$.