What is (x-1)^2+(y+3)^2-9=0 ?

The solution to (x-1)^2+(y+3)^2-9=0 is Circle with (a,b)=(1,-3),r=3

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(x-1)^2+(y+3)^2-9=0

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

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

Solution steps

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

Rewrite

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 1)^{2} + (y + 3)^{2} = 5

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

x^{2} + (y + 5)^{2} = 25

(x - 3)^{2} + y^{2} = \frac{1}{9}

(x + 2)^{2} + y^{2} = 36

What is (x-1)^2+(y+3)^2-9=0 ?
The solution to (x-1)^2+(y+3)^2-9=0 is Circle with (a,b)=(1,-3),r=3