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)^2+(y+10)^2=4\iff [x-(\underline{-2})]^2+[y-(\underline{-10})]^2=\underline{2}^2[/tex]
First, rewrite the expression in the form of a standard circle equation.
[tex]\left(x-\left(-2\right)\right)^2+\left(y-\left(-10\right)\right)^2=2^2[/tex]
We have our answer:
Center of circle: [tex]\left(-2,\:-10\right)[/tex]
[tex]radius=2[/tex]

Have an nice day! :)

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