What is x^2+y^2+12x-8y+3=0 ?

The solution to x^2+y^2+12x-8y+3=0 is Circle with (a,b)=(-6,4),r=7

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x^2+y^2+12x-8y+3=0

x^{2} + y^{2} + 12 x - 8 y + 3 = 0

(a, b) = (- 6, 4), r = 7

Solution steps

x^{2} + y^{2} + 12 x - 8 y + 3 = 0

Rewrite

x^{2} + y^{2} + 12 x - 8 y + 3 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - 4)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 6, 4), r = 7

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

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

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

y^{2} + z^{2} = 1

x^{2} + y^{2} + 4 x + 3 = 0

What is x^2+y^2+12x-8y+3=0 ?
The solution to x^2+y^2+12x-8y+3=0 is Circle with (a,b)=(-6,4),r=7