Respuesta :

19allenethm

Answer:

A: [tex](x-0)^{2} +(y-0)^{2} = 2^{2}[/tex]

or [tex]x^{2} +y^{2} =4[/tex]

B: [tex](x-5)^{2} +(y+4)^{2} =9[/tex]

Check the picture below.

[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} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{small}{circle}~~ \begin{cases} r=2\\ h=0\\ k=0 \end{cases}\implies (x-0)^2+(y-0)^2=2^2\implies x^2+y^2=4 \\\\\\ \stackrel{large}{circle}~~ \begin{cases} r=3\\ h=5\\ k=-4 \end{cases} \\[3em] [x-5]^2+[y-(-4)]^2=3^2\implies (x-5)^2+(y+4)^2=9[/tex]

Ver imagen jdoe0001

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