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center x^2+(y+3)^2=16

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center x²+(y+3)²=16

Solution

(0, −3)

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x²+(y+3)²=16

Circle Equation

(x−a)²+(y−b)²=r² is the circle equation with a radius r, centered at (a, b)

Rewrite x²+(y+3)²=16 in the form of the standard circle equation

(x−0)²+(y−(−3))²=4²

Therefore the circle properties are:

(a, b)=(0, −3), r=4

And the center is:

(0, −3)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

center x^2+(y+3)^2=16

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