What is 4x^2+4y^2+23x+25y-11=0 ?

The solution to 4x^2+4y^2+23x+25y-11=0 is Circle with (a,b)=(-23/8 ,-25/8),r=(sqrt(665))/(4sqrt(2))

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4x^2+4y^2+23x+25y-11=0

4 x^{2} + 4 y^{2} + 23 x + 25 y - 11 = 0

(a, b) = (- \frac{23}{8}, - \frac{25}{8}), r = \frac{665}{4 2}

Solution steps

4 x^{2} + 4 y^{2} + 23 x + 25 y - 11 = 0

Rewrite

4 x^{2} + 4 y^{2} + 23 x + 25 y - 11 = 0

in the form of the standard circle equation

(x - (- \frac{23}{8}))^{2} + (y - (- \frac{25}{8}))^{2} = (\frac{665}{4 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{23}{8}, - \frac{25}{8}), r = \frac{665}{4 2}

(x + 3)^{2} + (y - 3)^{2} = 36

(x + 10)^{2} + (y - 6)^{2} = 9

x^{2} + y^{2} + 4 x - 4 y - 28 = 0

x^{2} + y^{2} = 43.9569

x^{2} + y^{2} - 4 x - 45 = 0

What is 4x^2+4y^2+23x+25y-11=0 ?
The solution to 4x^2+4y^2+23x+25y-11=0 is Circle with (a,b)=(-23/8 ,-25/8),r=(sqrt(665))/(4sqrt(2))