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

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

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

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

(a, b) = (- 1, 5), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 5), r = 2

2 x^{2} + 3 x + 2 y^{2} + 7 y = 52.75

(x - 5)^{2} + (y - 2)^{2} = 49

(x - 15)^{2} + [y - (- 20)]^{2} = (9)^{2}

25 = (x + 7)^{2} + (y + 2)^{2}

x^{2} + y^{2} - 7 x + 2 y - 11 = 0

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