What is 16x^2+16y^2-48x-16y-41=0 ?

The solution to 16x^2+16y^2-48x-16y-41=0 is Circle with (a,b)=(3/2 , 1/2),r= 9/4

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16x^2+16y^2-48x-16y-41=0

16 x^{2} + 16 y^{2} - 48 x - 16 y - 41 = 0

(a, b) = (\frac{3}{2}, \frac{1}{2}), r = \frac{9}{4}

Solution steps

16 x^{2} + 16 y^{2} - 48 x - 16 y - 41 = 0

Rewrite

16 x^{2} + 16 y^{2} - 48 x - 16 y - 41 = 0

in the form of the standard circle equation

(x - \frac{3}{2})^{2} + (y - \frac{1}{2})^{2} = (\frac{9}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, \frac{1}{2}), r = \frac{9}{4}

(x - 2)^{2} + y^{2} = 10

x^{2} + y^{2} - 6 x + 10 y - 15 = 0

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

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

x^{2} + y^{2} = 53

What is 16x^2+16y^2-48x-16y-41=0 ?
The solution to 16x^2+16y^2-48x-16y-41=0 is Circle with (a,b)=(3/2 , 1/2),r= 9/4