Given points $A(0 ; 0), B(4 ; 0)$ and $C(0 ; 6)$. Formulate the equation of the circle circumscribed around triangle $A B C$.

#
Since the abscissas of points $A$ and $C$ are 0, these points lie on the line $x=0$, i.e., on the $O Y$ axis. Since the ordinates of points $A$ and $B$ are 0, these points lie on the line $y=0$, i.e., on the $O Y$ axis. Therefore, triangle $A B C$ is a right triangle, $\angle B A C=90^{\circ}$. Hence, the center of its circumscribed circle coincides with the midpoint $M\left(x_{0} ; y_{0}\right)$

of the hypotenuse $B C$, and the radius $R$ is half the hypotenuse.

Using the formulas for the coordinates of the midpoint of a segment, we find that

$$
x_{0}=\frac{4+0}{2}=2, y_{0}=\frac{0+6}{2}=3 .
$$

Using the formula for the distance between two points

$$
B C=\sqrt{(0-4)^{2}+(6-0)^{2}}=\sqrt{52}=2 \sqrt{13}
$$

Therefore, $R=\frac{1}{2} B C=\sqrt{13}$.

Thus, the desired equation is

$$
(x-2)^{2}+(y-3)^{2}=13
$$

## Answer

$(x-2)^{2}+(y-3)^{2}=13$