What is 4(x+3)^2+4(y-0)^2=16 ?

The solution to 4(x+3)^2+4(y-0)^2=16 is Circle with (a,b)=(-3,0),r=2

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4(x+3)^2+4(y-0)^2=16

4 (x + 3)^{2} + 4 (y - 0)^{2} = 16

(a, b) = (- 3, 0), r = 2

Solution steps

4 (x + 3)^{2} + 4 (y - 0)^{2} = 16

Rewrite

4 (x + 3)^{2} + 4 (y - 0)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 3, 0), r = 2

3 x^{2} + 3 y^{2} = 48

x^{2} + y^{2} + 2 x - 6 y - 40 = 0

x^{2} + y^{2} = \frac{9}{5}

x^{2} - 10 x + y^{2} - 20 y + 60 = 0

x^{2} + y^{2} + 8 x - 8 y - 4 = 0

What is 4(x+3)^2+4(y-0)^2=16 ?
The solution to 4(x+3)^2+4(y-0)^2=16 is Circle with (a,b)=(-3,0),r=2