What is 4x^2+4y^2-48x+20y=-141 ?

The solution to 4x^2+4y^2-48x+20y=-141 is Circle with (a,b)=(6,-5/2),r=sqrt(7)

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4x^2+4y^2-48x+20y=-141

4 x^{2} + 4 y^{2} - 48 x + 20 y = - 141

(a, b) = (6, - \frac{5}{2}), r = 7

Solution steps

4 x^{2} + 4 y^{2} - 48 x + 20 y = - 141

Rewrite

4 x^{2} + 4 y^{2} - 48 x + 20 y = - 141

in the form of the standard circle equation

(x - 6)^{2} + (y - (- \frac{5}{2}))^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (6, - \frac{5}{2}), r = 7

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

x^{2} + y^{2} + 10 x + 2 y + 22 = 0

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

(x - 5)^{2} + (y + 4)^{2} = 9

x^{2} + y^{2} + 10 x + 2 y = - 12

What is 4x^2+4y^2-48x+20y=-141 ?
The solution to 4x^2+4y^2-48x+20y=-141 is Circle with (a,b)=(6,-5/2),r=sqrt(7)