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

The solution to 4(x-3)^2+4y^2=4 is Circle with (a,b)=(3,0),r=1

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

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

(a, b) = (3, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 1

x^{2} + y^{2} + 2 x - 14 y = - 25

x^{2} + y^{2} + 4 x - 6 y - 16 = 0

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

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

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

What is 4(x-3)^2+4y^2=4 ?
The solution to 4(x-3)^2+4y^2=4 is Circle with (a,b)=(3,0),r=1