so, we know the center of this circle is at 6, -2, and a point on the circle is at 3,4. Now, from the center of the circle, to a point on the circle, is how long the radius is, therefore, the distance from 6, -2 to 3,4 is the radius "r", thus,
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ 6}} &,&{{ -2}}~)
% (c,d)
&&(~{{3}} &,&{{ 4}}~)
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
r=\sqrt{[3-6]^2+[4-(-2)]^2}\implies r=\sqrt{(3-6)^2+(4+2)^2}
\\\\\\
r=\sqrt{(-3)^2+6^2}\implies r=\sqrt{45}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \textit{equation of a circle}\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad
center~~(\stackrel{6}{{{ h}}},\stackrel{-2}{{{ k}}})\qquad \qquad
radius=\stackrel{\sqrt{45}}{{{ r}}}
\\\\\\\
[x-6]^2+[y-(-2)]^2=(\sqrt{45})^2\implies (x-6)^2+(y+2)^2=45[/tex]
