Respuesta :
[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}}
\end{array} [/tex]
Answer:
The center is at (4, -6), and the length of the radius is 5.
Step-by-step explanation:
(x − 4)2 + (y + 6)2 = 25
(x − 4)2 + (y − (-6))2 = 52
When I compare my equation with the standard form, (x − h)2 + (y − k)2 = r2, I get h = 4, k = -6, and r = 5. The center is at (4, -6), and the length of the radius is 5.
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