What is (x-5)^2+(y-3)^2=16 ?

The solution to (x-5)^2+(y-3)^2=16 is Circle with (a,b)=(5,3),r=4

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(x-5)^2+(y-3)^2=16

(x - 5)^{2} + (y - 3)^{2} = 16

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

Solution steps

(x - 5)^{2} + (y - 3)^{2} = 16

Rewrite

(x - 5)^{2} + (y - 3)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 1)^{2} + y^{2} = 6

2 x - 4 y - x^{2} - y^{2} - 1 \geq 0

(x - (- 15))^{2} + (y - 37)^{2} = 2^{2}

(x + 1)^{2} + (y + 3)^{2} = 25

(x - 2)^{2} + (y + 5)^{2} = 100

What is (x-5)^2+(y-3)^2=16 ?
The solution to (x-5)^2+(y-3)^2=16 is Circle with (a,b)=(5,3),r=4