What is (x+3)^2+(y-5)^2=4^2 ?

The solution to (x+3)^2+(y-5)^2=4^2 is Circle with (a,b)=(-3,5),r=4

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(x+3)^2+(y-5)^2=4^2

(x + 3)^{2} + (y - 5)^{2} = 4^{2}

(a, b) = (- 3, 5), r = 4

Solution steps

(x + 3)^{2} + (y - 5)^{2} = 4^{2}

Rewrite

(x + 3)^{2} + (y - 5)^{2} = 4^{2}

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 5)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 3, 5), r = 4

x^{2} + 5 x + y^{2} - 7 y + \frac{247}{16} = 0

(x - 5)^{(2)} + (y + 7)^{(2)} = 50

x^{2} - 16 x + 64 + y^{2} - 16 x = 0

x^{2} + y^{2} + 6 x - 4 y - 4 = 0

(x - 4)^{2} + (y + 3)^{2} = \frac{16}{9}

What is (x+3)^2+(y-5)^2=4^2 ?
The solution to (x+3)^2+(y-5)^2=4^2 is Circle with (a,b)=(-3,5),r=4