What is (x-3)^2+(y-5)^2-54=0 ?

The solution to (x-3)^2+(y-5)^2-54=0 is Circle with (a,b)=(3,5),r=sqrt(54)

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(x-3)^2+(y-5)^2-54=0

(x - 3)^{2} + (y - 5)^{2} - 54 = 0

(a, b) = (3, 5), r = 54

Solution steps

(x - 3)^{2} + (y - 5)^{2} - 54 = 0

Rewrite

(x - 3)^{2} + (y - 5)^{2} - 54 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 5)^{2} = (54)^{2}

Therefore the circle properties are:

(a, b) = (3, 5), r = 54

(x - (- 3))^{2} + (y - 0)^{2} = 6^{2}

(x - 6)^{(2)} + (y + 2)^{(2)} = 5

(x + 4)^{2} + (y + 1)^{2} = 30

x^{2} + y^{2} + 6 x - 8 y = - 24

y^{2} = 1 - 1 x^{2}

What is (x-3)^2+(y-5)^2-54=0 ?
The solution to (x-3)^2+(y-5)^2-54=0 is Circle with (a,b)=(3,5),r=sqrt(54)