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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

radius

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

r = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = \frac{3}{2}

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

(x - 17)^{2} + (y + 5)^{2} = 144

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

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

x^{2} + y^{2} - 8 x - 6 y - 11 = 0