What is (3x-2-2)^{(2)}+(3y+1)^{(2)}=9 ?

The solution to (3x-2-2)^{(2)}+(3y+1)^{(2)}=9 is Circle with (a,b)=(4/3 ,-1/3),r=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3x-2-2)^{(2)}+(3y+1)^{(2)}=9

(3 x - 2 - 2)^{(2)} + (3 y + 1)^{(2)} = 9

(a, b) = (\frac{4}{3}, - \frac{1}{3}), r = 1

Solution steps

(3 x - 2 - 2)^{2} + (3 y + 1)^{2} = 9

Rewrite

(3 x - 2 - 2)^{2} + (3 y + 1)^{2} = 9

in the form of the standard circle equation

(x - \frac{4}{3})^{2} + (y - (- \frac{1}{3}))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (\frac{4}{3}, - \frac{1}{3}), r = 1

x^{2} + y^{2} - 8 x + 6 y = - 24

x^{2} + y^{2} - 16 x + 8 y + 64 = 0

4 x^{2} + 4 y^{2} - 6 x + 10 y = \frac{1}{2}

(x + 15)^{2} + (y - 12)^{2} = 64

x^{2} + y^{2} = 1

What is (3x-2-2)^{(2)}+(3y+1)^{(2)}=9 ?
The solution to (3x-2-2)^{(2)}+(3y+1)^{(2)}=9 is Circle with (a,b)=(4/3 ,-1/3),r=1