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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 5)^{2} + (y + 3)^{2} = 4

x^{2} + y^{2} - 16 x = 0

x^{2} + y^{2} = 11 + 62

x^{2} + y^{2} - 8 x + 16 y + 16 = 0

(x - 1)^{2} + (y - 3)^{2} = 5^{2}

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