What is 16x^2+16y^2+16x+40y-7=0 ?

The solution to 16x^2+16y^2+16x+40y-7=0 is Circle with (a,b)=(-1/2 ,-5/4),r= 3/2

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16x^2+16y^2+16x+40y-7=0

16 x^{2} + 16 y^{2} + 16 x + 40 y - 7 = 0

(a, b) = (- \frac{1}{2}, - \frac{5}{4}), r = \frac{3}{2}

Solution steps

16 x^{2} + 16 y^{2} + 16 x + 40 y - 7 = 0

Rewrite

16 x^{2} + 16 y^{2} + 16 x + 40 y - 7 = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - \frac{5}{4}), r = \frac{3}{2}

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

x^{2} + (y - 5)^{2} = 9

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

x^{2} + y^{2} + 4 x - 10 y = 4

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

What is 16x^2+16y^2+16x+40y-7=0 ?
The solution to 16x^2+16y^2+16x+40y-7=0 is Circle with (a,b)=(-1/2 ,-5/4),r= 3/2