What is (x-6)^2+(y-3)^2=sqrt(20)^2 ?

The solution to (x-6)^2+(y-3)^2=sqrt(20)^2 is Circle with (a,b)=(6,3),r=sqrt((\sqrt{20))^2}

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(x-6)^2+(y-3)^2=sqrt(20)^2

(x - 6)^{2} + (y - 3)^{2} = 20^{2}

(a, b) = (6, 3), r = (20)^{2}

Solution steps

(x - 6)^{2} + (y - 3)^{2} = (20)^{2}

Rewrite

(x - 6)^{2} + (y - 3)^{2} = 20^{2}

in the form of the standard circle equation

(x - 6)^{2} + (y - 3)^{2} = ((20)^{2})^{2}

Therefore the circle properties are:

(a, b) = (6, 3), r = (20)^{2}

(x - 13)^{2} + (y + 7)^{2} = 144

x^{2} + (y - \frac{1}{4})^{2} = \frac{16}{9}

3 x^{2} + 3 y^{2} - 30 x + 36 y - 9 = 0

x^{2} + y^{2} + 2 x + 16 y + 61 = 0

x^{2} + y^{2} + 8 x + 18 y + 72 = 0

What is (x-6)^2+(y-3)^2=sqrt(20)^2 ?
The solution to (x-6)^2+(y-3)^2=sqrt(20)^2 is Circle with (a,b)=(6,3),r=sqrt((\sqrt{20))^2}