Respuesta :
[tex]\bf \textit{equation of a circle}\\\\
(x- h)^2+(y- k)^2= r^2
\qquad
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x-\stackrel{h}{2})^2+(y-\stackrel{k}{8})^2=16\implies (x-2)^2+(y-8)^2=\stackrel{r}{4^2}[/tex]
The radius is 4.
That equation is in center-radius form. When that’s the case, the radius is the square root of the number in the right of the equals sign. Square root of 16 is 4.
That equation is in center-radius form. When that’s the case, the radius is the square root of the number in the right of the equals sign. Square root of 16 is 4.
