$\left[\begin{array}{l}\text { Coordinate Method on the Plane }] \\ {[\quad \text { Circles (other). }}\end{array}\right]$

A circle with its center at point $M(3 ; 1)$ passes through the origin. Formulate the equation of the circle.
A circle with radius $R$ and center at point $A(a ; b)$ has the equation

$$
(x-a)^{2}+(y-b)^{2}=R^{2}
$$

In our case, $a=3, b=1$. Since the point $O(0 ; 0)$ lies on the circle $(x-3)^{2}+(y-1)^{2}=R^{2}$, the coordinates of this point satisfy the equation of the circle. From the equality $(0-3)^{2}+(0-1)^{2}=R^{2}$, we find that $R^{2}=9+1=$ 10.

## Answer

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