What is (x-4)^2+(y-1)^2=(sqrt(53))^2 ?

The solution to (x-4)^2+(y-1)^2=(sqrt(53))^2 is Circle with (a,b)=(4,1),r=sqrt((\sqrt{53))^2}

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(x-4)^2+(y-1)^2=(sqrt(53))^2

(x - 4)^{2} + (y - 1)^{2} = (53)^{2}

(a, b) = (4, 1), r = (53)^{2}

Solution steps

(x - 4)^{2} + (y - 1)^{2} = (53)^{2}

Rewrite

(x - 4)^{2} + (y - 1)^{2} = 53^{2}

in the form of the standard circle equation

(x - 4)^{2} + (y - 1)^{2} = ((53)^{2})^{2}

Therefore the circle properties are:

(a, b) = (4, 1), r = (53)^{2}

(x - 1)^{2} + (y - 3)^{2} = 8

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

8^{2} = x^{2} + y^{2}

x^{2} + y^{2} = 54

X^{2} + Y^{2} - X = 0

What is (x-4)^2+(y-1)^2=(sqrt(53))^2 ?
The solution to (x-4)^2+(y-1)^2=(sqrt(53))^2 is Circle with (a,b)=(4,1),r=sqrt((\sqrt{53))^2}