Given:
[tex](x-5)^2+y^2=16[/tex]
General equation of line is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
So the radius and center of circle is:
[tex]\begin{gathered} (h,k)=\text{ Point of circle} \\ r=\text{radius} \end{gathered}[/tex]
Compare the general equation so:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-5)^2+y^2=16 \\ (x-5)^2+y^2=(4)^2 \\ so \\ h=5 \\ k=0 \\ r=4 \end{gathered}[/tex]
so center of circle is:
[tex]\begin{gathered} (h,k)=(5,0) \\ \text{radius(r) }=4 \end{gathered}[/tex]
