What is x^2+y^2-10x-5y-40=0 ?

The solution to x^2+y^2-10x-5y-40=0 is Circle with (a,b)=(5, 5/2),r=(sqrt(285))/2

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x^2+y^2-10x-5y-40=0

x^{2} + y^{2} - 10 x - 5 y - 40 = 0

(a, b) = (5, \frac{5}{2}), r = \frac{285}{2}

Solution steps

x^{2} + y^{2} - 10 x - 5 y - 40 = 0

Rewrite

x^{2} + y^{2} - 10 x - 5 y - 40 = 0

in the form of the standard circle equation

(x - 5)^{2} + (y - \frac{5}{2})^{2} = (\frac{285}{2})^{2}

Therefore the circle properties are:

(a, b) = (5, \frac{5}{2}), r = \frac{285}{2}

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

2 x^{2} + 2 y^{2} - 4 x - 16 = 0

(x - 3)^{2} + (y - 0)^{2} = 4

(x + 5)^{2} + (y + 4)^{2} = 64

2 x^{2} + 2 y^{2} + 6 x + 10 y + 9 = 0

What is x^2+y^2-10x-5y-40=0 ?
The solution to x^2+y^2-10x-5y-40=0 is Circle with (a,b)=(5, 5/2),r=(sqrt(285))/2