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.
General eq. of circle is 

$(x-a)^2+(y-b)^2=r^2$ where r is radius & $(a,b)$ is the center.

Hint:Put value of $r=5$ then

$(x-a)^2+(y-b)^2=25$ 

Now put points which are situated on circle in place of (x,y) one by one you get two eq. solve them for values of $a$ & $b$ then you get the eq. of circle by using these values in general eq.