What is (x-4)^2+(y+1)^2=9 ?

The solution to (x-4)^2+(y+1)^2=9 is Circle with (a,b)=(4,-1),r=3

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(x-4)^2+(y+1)^2=9

(x - 4)^{2} + (y + 1)^{2} = 9

(a, b) = (4, - 1), r = 3

Solution steps

(x - 4)^{2} + (y + 1)^{2} = 9

Rewrite

(x - 4)^{2} + (y + 1)^{2} = 9

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 1))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (4, - 1), r = 3

(x + 5)^{2} + (y - 8)^{2} = 121

5 x^{2} + 5 y^{2} - 7 x + 11 y - 9 = 0

x^{2} + 10 x + y^{2} - 12 y = 3

x^{2} + y^{2} - 22 y - 5 = 0

x^{2} + y^{2} + 6 x + 8 y - 75 = 0

What is (x-4)^2+(y+1)^2=9 ?
The solution to (x-4)^2+(y+1)^2=9 is Circle with (a,b)=(4,-1),r=3