Hello, archistrategos214!

There's surely an easier way to do #5 . . . but here's my solution.

5) Find the equation of a circle tangent to the line $\displaystyle x+y\:=\:8$ at the point $\displaystyle (2,6)$

and passing through the point $\displaystyle (4,0)$. Code:

\ |
* A
| \(2,6)
| *
| / \
/ \
/ | \
/ | \
C * | \
- - - - - + - - - * - - - * - -
| (4,0) \
| B

The center of the circle is $\displaystyle C$.

Since the circle is tangent to the line at point $\displaystyle A$,

. . the radius at $\displaystyle A$ is perpendicular to the line.

The center $\displaystyle C$ lies along this perpendicular.

The line is: .$\displaystyle y \:=\:-x + 8$ with slope $\displaystyle -1$.

The equation of the perpendicular is: .$\displaystyle y - 6 \:=\:1(x - 2)\quad\Rightarrow\quad y\:=\:x + 4$ **[1]**

The center $\displaystyle C$ also lies on the perpendicular bisector of chord $\displaystyle AB$.

The midpoint of $\displaystyle AB$ is: $\displaystyle (3,3)$

The slope of $\displaystyle AB\:=\:\frac{0-6}{4-2}\:=\:-3$

. . The slope of the perpendicular bisector is: $\displaystyle +\frac{1}{3}$

The equation of the bisector is: .$\displaystyle y - 3 \:=\:\frac{1}{3}(x - 3)\quad\Rightarrow\quad y \:=\:\frac{1}{3}x + 2$ **[2]**

The center is the intersection of **[1]** and **[2]**: .$\displaystyle x + 4 \:=\:\frac{1}{3}x + 2\quad\Rightarrow\quad x = -3,\;y = 1$

. . Hence, the center is: $\displaystyle C(-3,1)$

The radius is the length of $\displaystyle CA:\;\;r^2 \:=\:(2+3)^2 + (6-1)^2 \:=\:50$

Therefore, the equation of the circle is: .$\displaystyle \boxed{(x+3)^2 + (y-1)^2\:=\:50}$