What is 4x^2+4y^2+20x-32y=5 ?

The solution to 4x^2+4y^2+20x-32y=5 is Circle with (a,b)=(-5/2 ,4),r=sqrt(47/2)

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

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

(a, b) = (- \frac{5}{2}, 4), r = \frac{47}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - 4)^{2} = (\frac{47}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 4), r = \frac{47}{2}

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

(x + 4)^{2} + (y + 2)^{2} = 25

What is 4x^2+4y^2+20x-32y=5 ?
The solution to 4x^2+4y^2+20x-32y=5 is Circle with (a,b)=(-5/2 ,4),r=sqrt(47/2)