The Solution.
The given equation of a circle is
[tex](x-1)^2+(y-2)^2=16\ldots eqn(1)[/tex]
By formula, the equation of a circle is
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2\ldots eqn(2) \\ \text{where the center is (a,b), and r=radius} \end{gathered}[/tex]
Comparing the corresponding terms in eqn(1) and eqn(2), we have
[tex]\begin{gathered} x-a=x-1 \\ \Rightarrow a=1 \end{gathered}[/tex]
Similarly,
[tex]\begin{gathered} y-b=y-2 \\ \Rightarrow b=2 \end{gathered}[/tex]
Also,
[tex]\begin{gathered} r^2=16 \\ r=\sqrt[]{16}=\pm4 \\ r=\text{ +4 units ( -4 is discarded)} \end{gathered}[/tex]
Hence, the center of the circle is (1,2) while the radius of the circle is 4 units.
