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}\\\\ -----------------------------\\\\ x^2+y^2=49\iff (x-0)^2+(y-0)^2=7^2\quad \begin{cases} h=0\\ k=0 \end{cases} \\\\\\ (x-\underline{(-3)})^2+(y-\underline{(+4)})^2=7^2\quad \begin{cases} h=-3\\ k=4 \end{cases} \\\\\\ \boxed{(x+3)^2+(y-4)^2=49}[/tex]
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The answer is:

[tex](x+3)^2+(y-4)^2=49[/tex]

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