What is 2x^2+2y^2-3x+5y-4=0 ?

The solution to 2x^2+2y^2-3x+5y-4=0 is Circle with (a,b)=(3/4 ,-5/4),r=(sqrt(33))/(2sqrt(2))

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2x^2+2y^2-3x+5y-4=0

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{33}{2 2}

Solution steps

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{33}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{33}{2 2}

(x - 3)^{2} + (y - 5)^{2} = 25

x^{2} + y^{2} + 10 x - 16 y + 64 = 0

2 (x^{2} - 3) + 2 (y^{2} - 1) = 45

(x + 2)^{2} + (y - 1.5)^{2} = (2.75 i)

(x + 2)^{2} + (y + 1)^{2} = 2

What is 2x^2+2y^2-3x+5y-4=0 ?
The solution to 2x^2+2y^2-3x+5y-4=0 is Circle with (a,b)=(3/4 ,-5/4),r=(sqrt(33))/(2sqrt(2))