What is 4x^2+28x+4y^2-8y=-17 ?

The solution to 4x^2+28x+4y^2-8y=-17 is Circle with (a,b)=(-7/2 ,1),r=3

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4x^2+28x+4y^2-8y=-17

4 x^{2} + 28 x + 4 y^{2} - 8 y = - 17

(a, b) = (- \frac{7}{2}, 1), r = 3

Solution steps

4 x^{2} + 28 x + 4 y^{2} - 8 y = - 17

Rewrite

4 x^{2} + 28 x + 4 y^{2} - 8 y = - 17

in the form of the standard circle equation

(x - (- \frac{7}{2}))^{2} + (y - 1)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, 1), r = 3

x^{2} + y^{2} \leq 2 x + 1

200 = x^{2} + y^{2} - 12 y + 36

4 x^{2} + 4 y^{2} - 24 x + 4 y + 17 = 0

x^{2} + y^{2} - 2 x + 4 y = 44

x^{2} + 4 x + y^{2} + 6 x = 0

What is 4x^2+28x+4y^2-8y=-17 ?
The solution to 4x^2+28x+4y^2-8y=-17 is Circle with (a,b)=(-7/2 ,1),r=3