What is 4x^2+4y^2+24x-16y-52=0 ?

The solution to 4x^2+4y^2+24x-16y-52=0 is Circle with (a,b)=(-3,2),r=sqrt(26)

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4x^2+4y^2+24x-16y-52=0

4 x^{2} + 4 y^{2} + 24 x - 16 y - 52 = 0

(a, b) = (- 3, 2), r = 26

Solution steps

4 x^{2} + 4 y^{2} + 24 x - 16 y - 52 = 0

Rewrite

4 x^{2} + 4 y^{2} + 24 x - 16 y - 52 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 2)^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 26

4 x^{2} + 8 x + 4 y^{2} + 4 = 8

x^{2} + y^{2} + 8 x - 14 y + 49 = 0

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

x^{2} + y^{2} - 6 x + 5 = 5

x^{2} + y^{2} - 6 x + 5 = 3

What is 4x^2+4y^2+24x-16y-52=0 ?
The solution to 4x^2+4y^2+24x-16y-52=0 is Circle with (a,b)=(-3,2),r=sqrt(26)