What is 4(x+1)^2+4(y)^2=121 ?

The solution to 4(x+1)^2+4(y)^2=121 is Circle with (a,b)=(-1,0),r= 11/2

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4(x+1)^2+4(y)^2=121

4 (x + 1)^{2} + 4 (y)^{2} = 121

(a, b) = (- 1, 0), r = \frac{11}{2}

Solution steps

4 (x + 1)^{2} + 4 y^{2} = 121

Rewrite

4 (x + 1)^{2} + 4 y^{2} = 121

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 0)^{2} = (\frac{11}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 1, 0), r = \frac{11}{2}

(x - 5)^{2} + (y + 3)^{2} = 73

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

(x - 5)^{2} + (y + 3)^{2} = 85

(x - 1)^{2} + (y - 4)^{2} = 12

x^{2} + (y + 1)^{2} - 5 = 0

What is 4(x+1)^2+4(y)^2=121 ?
The solution to 4(x+1)^2+4(y)^2=121 is Circle with (a,b)=(-1,0),r= 11/2