5.21. Form the equation of the circle circumscribed around the triangle formed by the lines $y=0.2 x-0.4, y=x+2, y=8-x$.
5.21. The angular coefficients of the lines $y=x+2$ and $y=8-x$ are $k_{1}=1$ and $k_{2}=-1$, respectively. Since $k_{1} k_{2} = -1$, the condition (5.9) for perpendicularity of the lines is satisfied; hence, $\triangle A B C$ is a right triangle (Fig. 5.11) and the center of the circle is the midpoint of its hypotenuse $A B$. Let's find the points of intersection of the line $y=0.2 x-0.4$ with the lines $y=x+2$ and $y=8-x$; solving the systems of equations

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-606.jpg?height=454&width=431&top_left_y=694&top_left_x=721)

Fig. 5.11

$$
\left\{\begin{array}{l}
y=0.2 x-0.4, \\
y=x+2
\end{array} \text { and } \left\{\begin{array}{l}
y=0.2 x-0.4 \\
y=8-x
\end{array}\right.\right.
$$

we obtain the points $A(-3 ;-1)$ and $B(7 ; 1)$ - the endpoints of the hypotenuse. Using formulas (5.2), we find the coordinates of the center of the circle: $O_{1}(2 ; 0)$. By formula (5.1), the radius of the circle is $R=O_{1} A=\sqrt{(-3-2)^{2}+(1-0)^{2}}=\sqrt{26}$. Finally, according to formula (5.11), we obtain the desired equation of the circle: $(x-2)^{2}+y^{2}=26$.

Answer: $(x-2)^{2}+y^{2}=26$.