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

The radius 2x^2+2y^2-3x=0 is r= 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

radius

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

r = \frac{3}{4}

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 radius is:

r = \frac{3}{4}

x^{2} - 8 x + y^{2} - 10 y = 40

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

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

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

x^{2} + y^{2} - 16 x + 12 y + 64 = 0