What is 4x^2+4y^2-16x+24y-51=0 ?

The solution to 4x^2+4y^2-16x+24y-51=0 is Circle with (a,b)=(2,-3),r=(sqrt(103))/2

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4x^2+4y^2-16x+24y-51=0

4 x^{2} + 4 y^{2} - 16 x + 24 y - 51 = 0

(a, b) = (2, - 3), r = \frac{103}{2}

Solution steps

4 x^{2} + 4 y^{2} - 16 x + 24 y - 51 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x + 24 y - 51 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = (\frac{103}{2})^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = \frac{103}{2}

(x + 1)^{2} + (y - 1)^{2} = 40

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

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

(x + 1)^{2} + (y - 1)^{2} = 36

x^{2} + y^{2} + 6 x + 10 y + 9 = 0

What is 4x^2+4y^2-16x+24y-51=0 ?
The solution to 4x^2+4y^2-16x+24y-51=0 is Circle with (a,b)=(2,-3),r=(sqrt(103))/2