What is (x-3)^2+(y+2)^2=4^2 ?

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

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(x-3)^2+(y+2)^2=4^2

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

(a, b) = (3, - 2), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 2), r = 4

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

x^{2} + y^{2} + 4 x - 10 = 0

x^{2} + y^{2} - 4 x + 12 y - 20 = 0

x^{2} + y^{2} + 10 x - 6 y + 9 = 0

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

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