What is x^2+y^2+10x-8y+37=0 ?

The solution to x^2+y^2+10x-8y+37=0 is Circle with (a,b)=(-5,4),r=2

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x^2+y^2+10x-8y+37=0

x^{2} + y^{2} + 10 x - 8 y + 37 = 0

(a, b) = (- 5, 4), r = 2

Solution steps

x^{2} + y^{2} + 10 x - 8 y + 37 = 0

Rewrite

x^{2} + y^{2} + 10 x - 8 y + 37 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, 4), r = 2

(x - 5)^{2} + (y + 6)^{2} = 20

\frac{x ^{2}}{16} + \frac{y ^{2}}{16} = 1

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

x^{2} + y^{2} - 10 x + 2 y + 22 = 0

(x - \frac{1}{3})^{2} + (y - 2)^{2} = 81

What is x^2+y^2+10x-8y+37=0 ?
The solution to x^2+y^2+10x-8y+37=0 is Circle with (a,b)=(-5,4),r=2