so hmm notice the picture below
a)
the center of the circle is the midpoint of those two folks
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -10}}\quad ,&{{ -2}})\quad
% (c,d)
&({{ 4}}\quad ,&{{ 6}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{4-10}{2}\quad ,\quad \cfrac{6-2}{2} \right)\impliedby \textit{center of the circle}[/tex]
b)
the diameter is the distance between P and Q, or the length of that segment, and the radius is half the diameter
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -10}}\quad ,&{{ -2}})\quad
% (c,d)
&({{ 4}}\quad ,&{{ 6}})
\end{array}\quad
% distance value
d = \sqrt{({{ 4}}-{{ (-10)}})^2 + ({{ 6}}-{{ (-2)}})^2}
\\\\\\
d=\cfrac{\sqrt{(4+10)^2+(6+2)^2}}{2}\impliedby \textit{radius of the circle}[/tex]
c)
so, from a), you found the h,k coordinates for the center, from b) you've got the radius
so, just plug them in here then [tex]\bf \textit{equation of a circle}\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad
radius=&{{ r}}
\end{array} [/tex]
