What is the center 2x^2+2y^2-3x=0 ?

The center 2x^2+2y^2-3x=0 is (3/4 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 2x^2+2y^2-3x=0

center

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

(\frac{3}{4}, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, 0), r = \frac{3}{4}

And the center is:

(\frac{3}{4}, 0)

x^{2} + y^{2} - 12 y + 32 = 0

x^{2} + y^{2} - 25 = 0

(x + 2)^{2} + y^{2} = 25

144 x^{2} + 144 y^{2} + 72 x + 192 y + 37 = 0

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5