Respuesta :

Check the picture below.

[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{0}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{2}{ r} \\\\[-0.35em] ~\dotfill\\[1em] [x-0]^2+[y-(-3)]^2=2^2\implies x^2+(y+3)^2=4[/tex]

Ver imagen jdoe0001

Answer:

x^2  + (y + 3)^2 = 4

or

x^2 + y^2 + 6y + 5 = 0.

Step-by-step explanation:

We see from the diagram that the center of the circle is at (0, -3)  and radius is 2.

The general form for a circle is  (x - a)^2 + (y - b)^2 = r^2  where  (a, b) is the center and r = the radius.

So substituting the given values the equation is:

(x - 0)^2 + (y - -3)^2 = 2^2

x^2  + (y + 3)^2 = 4  (answer).

If you want it in expanded form it is

x^2 + y^2 + 6y + 5 = 0.

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