What is 1/2 x^2+1/2 y^2+5/2 x+10y+5=0 ?

The solution to 1/2 x^2+1/2 y^2+5/2 x+10y+5=0 is Circle with (a,b)=(-5/2 ,-10),r=(sqrt(385))/2

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1/2 x^2+1/2 y^2+5/2 x+10y+5=0

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} + \frac{5}{2} x + 10 y + 5 = 0

(a, b) = (- \frac{5}{2}, - 10), r = \frac{385}{2}

Solution steps

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} + \frac{5}{2} x + 10 y + 5 = 0

Rewrite

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} + \frac{5}{2} x + 10 y + 5 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - (- 10))^{2} = (\frac{385}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, - 10), r = \frac{385}{2}

(x - 7)^{2} + (y + 2)^{2} = 100

x^{2} + y^{2} - 14 x - 4 y + 17 = 0

(x - 2)^{2} + (y - 4)^{2} = 17

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

x^{2} + y^{2} + 4 x - 6 y + 10 = 0

What is 1/2 x^2+1/2 y^2+5/2 x+10y+5=0 ?
The solution to 1/2 x^2+1/2 y^2+5/2 x+10y+5=0 is Circle with (a,b)=(-5/2 ,-10),r=(sqrt(385))/2