What is x^2+y^2+8x+4y-3=40 ?

The solution to x^2+y^2+8x+4y-3=40 is Circle with (a,b)=(-4,-2),r=3sqrt(7)

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x^2+y^2+8x+4y-3=40

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

(a, b) = (- 4, - 2), r = 37

Solution steps

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

Rewrite

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 2))^{2} = (37)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 2), r = 37

y^{2} = - x

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

\frac{x ^{2}}{25} + \frac{y ^{2}}{25} = 1

x^{2} + y^{2} - 2 x + 4 y - 31 = 0

y^{2} = - 36 x

What is x^2+y^2+8x+4y-3=40 ?
The solution to x^2+y^2+8x+4y-3=40 is Circle with (a,b)=(-4,-2),r=3sqrt(7)