What is 4x^2+4y^2+24x-8y-41=0 ?

The solution to 4x^2+4y^2+24x-8y-41=0 is Circle with (a,b)=(-3,1),r= 9/2

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4x^2+4y^2+24x-8y-41=0

4 x^{2} + 4 y^{2} + 24 x - 8 y - 41 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 24 x - 8 y - 41 = 0

Rewrite

4 x^{2} + 4 y^{2} + 24 x - 8 y - 41 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 6 x - 12 y + 36 = 0

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

(x - 2)^{(2)} + (y + 1)^{(2)} \geq 4

(x - 2)^{2} + (y - 6)^{2} = 16

(x - 3)^{2} + (y - 5)^{2} = 6^{2}

What is 4x^2+4y^2+24x-8y-41=0 ?
The solution to 4x^2+4y^2+24x-8y-41=0 is Circle with (a,b)=(-3,1),r= 9/2