8,9

Form the equation of a circle that touches the coordinate axes and passes through the point $A(2 ; 1)$.

#
Since the circle touches the coordinate axes and passes through a point located in the first coordinate quadrant, the center of the circle lies on the line $y=x$. Therefore, the abscissa and ordinate of the center of the circle are equal to its radius. Thus, the equation of the circle has the form

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

Since the point $A(2 ; 1)$ lies on the circle, the coordinates of this point satisfy the obtained equation, i.e., $(2-R)^{2}+(1-R)^{2}=R^{2}$. From this, we find that $R=1$ or $R=5$. Therefore, the desired equation has the form:

$$
(x-5)^{2}+(y-5)^{2}=25 \text { or }(x-1)^{2}+(y-1)^{2}=1 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-43.jpg?height=675&width=818&top_left_y=0&top_left_x=623)

## Answer

$(x-5)^{2}+(y-5)^{2}=25$ or $(x-1)^{2}+(y-1)^{2}=1$.