What is (x-20)^2+(y-0.05)^2=81 ?

The solution to (x-20)^2+(y-0.05)^2=81 is Circle with (a,b)=(20,0.05),r=9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-20)^2+(y-0.05)^2=81

(x - 20)^{2} + (y - 0.05)^{2} = 81

(a, b) = (20, 0.05), r = 9

Solution steps

(x - 20)^{2} + (y - 0.05)^{2} = 81

Rewrite

(x - 20)^{2} + (y - 0.05)^{2} = 81

in the form of the standard circle equation

(x - 20)^{2} + (y - 0.05)^{2} = 9^{2}

Therefore the circle properties are:

(a, b) = (20, 0.05), r = 9

2 x^{2} + 2 y^{2} - 2 x + 18 y + 33 = 0

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

(x - 9)^{2} + (y + 1)^{2} = 36

9 x^{2} + 9 y^{2} = 4

x^{2} + y^{2} = \frac{4}{9}

What is (x-20)^2+(y-0.05)^2=81 ?
The solution to (x-20)^2+(y-0.05)^2=81 is Circle with (a,b)=(20,0.05),r=9