What is x^2+y^2-10x+4y+20=0 ?

The solution to x^2+y^2-10x+4y+20=0 is Circle with (a,b)=(5,-2),r=3

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x^2+y^2-10x+4y+20=0

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

(a, b) = (5, - 2), r = 3

Solution steps

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

Rewrite

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, - 2), r = 3

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

(x - 3)^{2} + (y - 5)^{2} = 16

x^{2} + y^{2} = \frac{25}{36}

(x - 3)^{2} + (y + 4)^{2} = 9

x^{2} + y^{2} + 4 x + 12 y - 100 = 0

What is x^2+y^2-10x+4y+20=0 ?
The solution to x^2+y^2-10x+4y+20=0 is Circle with (a,b)=(5,-2),r=3