What is ((x+5)^2)/4+((y)^2)/4 =1 ?

The solution to ((x+5)^2)/4+((y)^2)/4 =1 is Circle with (a,b)=(-5,0),r=2

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((x+5)^2)/4+((y)^2)/4 =1

\frac{( x + 5 ) ^{2}}{4} + \frac{( y ) ^{2}}{4} = 1

(a, b) = (- 5, 0), r = 2

Solution steps

\frac{( x + 5 ) ^{2}}{4} + \frac{y ^{2}}{4} = 1

Rewrite

\frac{( x + 5 ) ^{2}}{4} + \frac{y ^{2}}{4} = 1

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- 5, 0), r = 2

(x - 2)^{2} + y^{2} = 100

x^{2} - 4 x + y^{2} - 2 y = 11

x^{2} + y^{2} + 4 x + 8 y - 36 = 0

x^{2} + y^{2} - 6 x - 8 y - 22 = 0

x^{2} + y^{2} - 9 x - 2 y + 17 = 0

What is ((x+5)^2)/4+((y)^2)/4 =1 ?
The solution to ((x+5)^2)/4+((y)^2)/4 =1 is Circle with (a,b)=(-5,0),r=2