What is (x+1)^2+(y-5)^2=20 ?

The solution to (x+1)^2+(y-5)^2=20 is Circle with (a,b)=(-1,5),r=sqrt(20)

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(x+1)^2+(y-5)^2=20

(x + 1)^{2} + (y - 5)^{2} = 20

(a, b) = (- 1, 5), r = 20

Solution steps

(x + 1)^{2} + (y - 5)^{2} = 20

Rewrite

(x + 1)^{2} + (y - 5)^{2} = 20

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 5)^{2} = (20)^{2}

Therefore the circle properties are:

(a, b) = (- 1, 5), r = 20

x^{2} + y^{2} + 6 x - 8 y + 13 = 0

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

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

x^{2} - 6 x + y^{2} - 20 y + 44 = 0

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

What is (x+1)^2+(y-5)^2=20 ?
The solution to (x+1)^2+(y-5)^2=20 is Circle with (a,b)=(-1,5),r=sqrt(20)