What is (x-2)^2+(y+4)^2=41 ?

The solution to (x-2)^2+(y+4)^2=41 is Circle with (a,b)=(2,-4),r=sqrt(41)

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(x-2)^2+(y+4)^2=41

(x - 2)^{2} + (y + 4)^{2} = 41

(a, b) = (2, - 4), r = 41

Solution steps

(x - 2)^{2} + (y + 4)^{2} = 41

Rewrite

(x - 2)^{2} + (y + 4)^{2} = 41

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = (41)^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 41

y^{2} + x^{2} - 8 x + 18 y + 72 = 0

x^{2} + y^{2} + 8 x + 4 y + 19 = 0

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

x^{2} + 2 x + y^{2} - 2 y = - 1

x^{2} + y^{2} + 10 x - 23 = 0

What is (x-2)^2+(y+4)^2=41 ?
The solution to (x-2)^2+(y+4)^2=41 is Circle with (a,b)=(2,-4),r=sqrt(41)