9x^2+9y^2-6x+3y-1=0

Question

find the center, radius, and intercepts of the circle and then sketch the graph 

Answer 1

9x² + 9y² - 6x + 3y - 1 = 0 ---> 9x²/9 + 9y²/9 - 6x/9 + 3y/9 - 1/9 = 0 ---->
(x² - 2*(1/3)x + 1/9) - 1/9 + (y² + 2*(1/3*2)y + 1/36) - 1/36 - 1/9 = 0
             ↓                                           ↓  
      (x - 1/3)²                   +             (y + 1/6)²  - (1/9 + 1/36 + 1/9) = 0
                                                                                   ↓
                                                                            - 9/36 or -1/4
(x - 1/3)² + (y + 1/6)² = (1/2)²
Coordinates of the center: (1/3 , -1/6) , r = 1/2
x-intercept (x,0): 9x² - 6x - 1 = 0 ----> x₁ = (1+√2)/3 , x₂ = (1-√2)/3
y-intercept (0,y): 9y² + 3y - 1 = 0 --->  y₁ = (-1 + √5)/6 , y₂ = (-1 - √5)/6