Hello, archistrategos214!

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

Code:

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

The center of the circle is .

Since the circle is tangent to the line at point ,

. . the radius at is perpendicular to the line.

The center lies along this perpendicular.

The line is: . with slope .

The equation of the perpendicular is: . **[1]**

The center also lies on the perpendicular bisector of chord .

The midpoint of is:

The slope of

. . The slope of the perpendicular bisector is:

The equation of the bisector is: . **[2]**

The center is the intersection of **[1]** and **[2]**: .

. . Hence, the center is:

The radius is the length of

Therefore, the equation of the circle is: .