What is (2x^2+2y^2-3x+5y-3)/2 =0 ?

The solution to (2x^2+2y^2-3x+5y-3)/2 =0 is Circle with (a,b)=(3/4 ,-5/4),r=(sqrt(29))/(2sqrt(2))

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(2x^2+2y^2-3x+5y-3)/2 =0

\frac{2 x ^{2} + 2 y ^{2} - 3 x + 5 y - 3}{2} = 0

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{29}{2 2}

Solution steps

\frac{2 x ^{2} + 2 y ^{2} - 3 x + 5 y - 3}{2} = 0

Rewrite

\frac{2 x ^{2} + 2 y ^{2} - 3 x + 5 y - 3}{2} = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{29}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{29}{2 2}

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

6 x^{2} + 6 y^{2} - 24 x + 36 y = 72

x^{2} + y^{2} - 2 (x + y) > 0

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

x^{2} + y^{2} + 8 x - 6 y - 15 = 0

What is (2x^2+2y^2-3x+5y-3)/2 =0 ?
The solution to (2x^2+2y^2-3x+5y-3)/2 =0 is Circle with (a,b)=(3/4 ,-5/4),r=(sqrt(29))/(2sqrt(2))