Prove that the line $3 x-4 y+25=0$ is tangent to the circle $x^{2}+y^{2}=25$ and find the coordinates of the point of tangency.

## Answer

$(-3 ; 4)$.

Let's find the coordinates of all common points of the given line and circle. For this, we will solve the system of equations

$$
\left\{\begin{array}{l}
3 x-4 y+25=0 \\
x^{2}+y^{2}=25
\end{array}\right.
$$

We will find that the circle and the line have only one common point with coordinates $x_{0}=-3, y_{0}=4$.

Therefore, the line is tangent to the circle.

Formulate the equation of the circle that is tangent to 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: $(x-5)^{2}+(y-5)^{2}=25$ or $(x-1)^{2}+(y-1)^{2}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-19.jpg?height=683&width=809&top_left_y=799&top_left_x=630)

## Answer

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