What is the radius x^2+y^2-10x-8y+32=0 ?

The radius x^2+y^2-10x-8y+32=0 is r=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius x^2+y^2-10x-8y+32=0

radius

x^{2} + y^{2} - 10 x - 8 y + 32 = 0

r = 3

Solution steps

x^{2} + y^{2} - 10 x - 8 y + 32 = 0

Rewrite

x^{2} + y^{2} - 10 x - 8 y + 32 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, 4), r = 3

And the radius is:

r = 3

x^{2} + y^{2} + 6 x + 8 y + 19 = 0

x^{2} + 4 x + y^{2} - 6 y + 9 = 0

x^{2} + y^{2} - 8 x - 8 y + 28 = 0

4 x^{2} + 4 y^{2} + 40 x - 32 y = 0

x^{2} + y^{2} - 4 x + 6 y = 16