What is (x+1)^2+(y+6)^2=19 ?

The solution to (x+1)^2+(y+6)^2=19 is Circle with (a,b)=(-1,-6),r=sqrt(19)

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(x+1)^2+(y+6)^2=19

(x + 1)^{2} + (y + 6)^{2} = 19

(a, b) = (- 1, - 6), r = 19

Solution steps

(x + 1)^{2} + (y + 6)^{2} = 19

Rewrite

(x + 1)^{2} + (y + 6)^{2} = 19

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 6))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 6), r = 19

(x + 5)^{2} + (y - 2)^{2} = 16

x^{2} + y^{2} + 10 y - 75 = 0

9 = 2 y - y^{2} - 6 x - x^{2}

x^{2} + y^{2} + 4 x - 10 y = - 4

x^{2} + y^{2} - 6 x + 14 y + 22 = 0

What is (x+1)^2+(y+6)^2=19 ?
The solution to (x+1)^2+(y+6)^2=19 is Circle with (a,b)=(-1,-6),r=sqrt(19)