What is 4x^2+4y^2-32x+8y-32=0 ?

The solution to 4x^2+4y^2-32x+8y-32=0 is Circle with (a,b)=(4,-1),r=5

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4x^2+4y^2-32x+8y-32=0

4 x^{2} + 4 y^{2} - 32 x + 8 y - 32 = 0

(a, b) = (4, - 1), r = 5

Solution steps

4 x^{2} + 4 y^{2} - 32 x + 8 y - 32 = 0

Rewrite

4 x^{2} + 4 y^{2} - 32 x + 8 y - 32 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, - 1), r = 5

4 x^{2} + 4 y^{2} + 20 x - 32 y = 5

x^{2} + y^{2} - 20 x + 16 y + 128 = 0

25 - x^{2} - y^{2} \geq 0

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

\frac{( x - 3 ) ^{2} + y ^{2}}{4} = 1

What is 4x^2+4y^2-32x+8y-32=0 ?
The solution to 4x^2+4y^2-32x+8y-32=0 is Circle with (a,b)=(4,-1),r=5