[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}}\\
&\uparrow &\uparrow &\uparrow \\
&1&-2&4
\end{array}
\\\\\\
(x-1)+(y-(-2))=4^2\implies (x-1)+(y+2)=4^2\\\\
-----------------------------\\\\
\begin{cases}
x=3.4\\
y=1.2
\end{cases}\implies (3.4-1)+(1.2+2)=4^2[/tex]
so hmm does the left-hand-side give you 16?
if it does, then the equation is true, and that point does lie on the circle
