<s>[INST] Analytic Geometry How does one solve: 
Find the equation of the circle which has it's center on the line $y= 3-x$ , and which has as tangents the lines $ 2y-x = 22, $       $   2x+y=11 $     ? 
 [/INST] Let $(a,b)$ be the center of the circle. Since the center lies on the line $y=3-x$, then we have $b=3-a$. The equation of the circle if the center on $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2\tag1
$$
and the equation of its tangent line is
$$
(x_c-a)(x-a)+(y_c-b)(y-b)=r^2,\tag2
$$
where $(x_c,y_c)$ is point of contact. Equation $(2)$ can be written as
$$
(x_c-a)x+(y_c-b)y=r^2+a(x_c-a)+b(y_c-b).\tag3
$$
The tangent lines are
$$
-x+2y=22\tag4
$$
and
$$
2x+y=11.\tag5
$$
Using $b=3-a$, comparing $(3)$ and $(4)$ yields $x_1-a=-1$, $y_1-b=2$, and 
$$
\begin{align}
r^2+a(x_1-a)+b(y_1-b)&=22\\
r^2+a(-1)+b(2)&=22\\
r^2-a+2(3-a)&=22\\
r^2-3a&=16.\tag6
\end{align}
$$
Similarly, comparing $(3)$ and $(5)$ yields $x_2-a=2$, $y_2-b=1$, and 
$$
\begin{align}
r^2+a(x_2-a)+b(y_2-b)&=11\\
r^2+a&=8.\tag7
\end{align}
$$
Solving $(6)$ and $(7)$ yields $a=-2$, $b=5$, and $r^2=10$. Thus, using $(1)$, the equation of the circle is
$$
\Large\color{blue}{(x+2)^2+(y-5)^2=10}.
$$
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
</s>