What is 16x^2+80x+16y^2-112y+247=0 ?

The solution to 16x^2+80x+16y^2-112y+247=0 is Circle with (a,b)=(-5/2 , 7/2),r= 7/4

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16x^2+80x+16y^2-112y+247=0

16 x^{2} + 80 x + 16 y^{2} - 112 y + 247 = 0

(a, b) = (- \frac{5}{2}, \frac{7}{2}), r = \frac{7}{4}

Solution steps

16 x^{2} + 80 x + 16 y^{2} - 112 y + 247 = 0

Rewrite

16 x^{2} + 80 x + 16 y^{2} - 112 y + 247 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - \frac{7}{2})^{2} = (\frac{7}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, \frac{7}{2}), r = \frac{7}{4}

(x + 3)^{2} + (y - 4)^{2} = 25

x^{2} + y^{2} + 6 x - 2 y - 3 = 0

(x + 2)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} - x + 2 y + 1 = 0

2 x^{2} + 2 y^{2} + 8 x + 7 = 0

What is 16x^2+80x+16y^2-112y+247=0 ?
The solution to 16x^2+80x+16y^2-112y+247=0 is Circle with (a,b)=(-5/2 , 7/2),r= 7/4