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

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

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

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

(a, b) = (5, 4), r = (3)^{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, 4), r = (3)^{2}

x^{2} + y^{2} + 2 x - 10 y = 0

3 (x - 1)^{2} + 3 (y - 2)^{2} = 12

- x^{2} - 4 x = y^{2} - 20 - 2 y

3 x^{2} + 3 y^{2} = 12

x^{2} + y^{2} + 6 x - 8 y + 24 = 0

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