Question: Let $X = \{ (x,y) \in \mathbb{R}^2 : x > 0, y \geq 0 \} $. I am claiming $X$ is open, but it is not closed. 

My Try:

To show it is not closed, I found a sequence that converges to a point outside $X$. For instance, $(x\_n, y\_n) = (\frac{1}{n}, \frac{1}{n} ) \to (0,0) \notin X$. Hence, $X$ cannot be closed. 

In trying to show it is open, I have diffictulty trying to find a suitable open ball contained in $X$. Any help would be greatly appreaciated. 

Thanks.

Answer: 
The centre of the circle lies on the perpendicular bisector of the line segment that joins $A$ and $B$. 

The midpoint of this line segment is $(-6,-1)$. The line segment has slope $-\frac{1}{2}$, so the perpendicular bisector has slope $2$.

Now we can find the equation of the perpendicular bisector. It is $y=2x+11$.

The distance from $(x,2x+11)$ to $(-8,0)$ is $5$. Thus
$$(x+8)^2+(2x+11)^2=25.$$
This simplifies to 
$$5x^2+60x+160=0,$$
and then to $x^2+12x+32=0$. The roots of this equation are easy to find: we get lucky, the quadratic even factors nicely.
$$(8+x\_1)^2 + y\_1^2 = 5^2$$
$$(4+x\_1)^2 + (2+y\_1)^2 = 5^2$$
Subtracting the second equation from first:
$$4(12+2x\_1) - 2(2+2y\_1) = 0$$
$$11 + 2x\_1 = y\_1$$

Substitute this in one of the two equations and find $x\_1$ and then eventually $y\_1$.Once you get the center of the circle $(x\_1,y\_1)$, you can calculate the equation of the circle using:
$$(x-x\_1)^2 + (y-y\_1)^2 = r^2$$
where $r=5$ in this case.