What is x^2+y^2-4x+10y+13=0 ?

The solution to x^2+y^2-4x+10y+13=0 is Circle with (a,b)=(2,-5),r=4

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x^2+y^2-4x+10y+13=0

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

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

Solution steps

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

Rewrite

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

x = 3 y^{2}

3 y^{2} = 24 x

f (x) = x^{2} - 6 x

What is x^2+y^2-4x+10y+13=0 ?
The solution to x^2+y^2-4x+10y+13=0 is Circle with (a,b)=(2,-5),r=4