What is 4x^2+4y^2-20x-12y-110=0 ?

The solution to 4x^2+4y^2-20x-12y-110=0 is Circle with (a,b)=(5/2 , 3/2),r=6

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4x^2+4y^2-20x-12y-110=0

4 x^{2} + 4 y^{2} - 20 x - 12 y - 110 = 0

(a, b) = (\frac{5}{2}, \frac{3}{2}), r = 6

Solution steps

4 x^{2} + 4 y^{2} - 20 x - 12 y - 110 = 0

Rewrite

4 x^{2} + 4 y^{2} - 20 x - 12 y - 110 = 0

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - \frac{3}{2})^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, \frac{3}{2}), r = 6

a^{2} + b^{2} = 35

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

(x + 8)^{2} + (y + 9)^{2} = 256

x^{2} + y^{2} + 2 x - 6 y = 26

x^{2} + 3 x + y^{2} + 8 x = 25

What is 4x^2+4y^2-20x-12y-110=0 ?
The solution to 4x^2+4y^2-20x-12y-110=0 is Circle with (a,b)=(5/2 , 3/2),r=6