What is the radius x^2+y^2+10x+12y+25=0 ?

The radius x^2+y^2+10x+12y+25=0 is r=6

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+12y+25=0

radius

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

r = 6

Solution steps

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

Rewrite

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 6))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 6), r = 6

And the radius is:

r = 6

x^{2} + y^{2} - 10 x - 8 y = - 5

x^{2} - 2 x + y^{2} - 8 y + 1 = 0

x^{2} + y^{2} - 8 x + 12 y + 10 = 0

x^{2} + y^{2} - 4 x + 8 y + 5 = 0

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

What is the radius x^2+y^2+10x+12y+25=0 ?
The radius x^2+y^2+10x+12y+25=0 is r=6