# How do you write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0)?

Mar 19, 2018

The equation of circle is ${x}^{2} + {y}^{2} + 6 x = 0$

#### Explanation:

Let the equation of circle be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

Points $j \left(- 6 , 0\right) , k \left(- 3 , 3\right) , l \left(0 , 0\right)$

$\therefore {\left(- 6\right)}^{2} + {0}^{2} + 2 g \cdot \left(- 6\right) + 2 f \cdot 0 + c = 0$

:. (36-12g +c= 0 or -12g+36+c=0(1) , similarly

$\therefore {\left(- 3\right)}^{2} + {3}^{2} + 2 g \cdot \left(- 3\right) + 2 f \cdot 3 + c = 0$

:. (9+9-6g +6f +c= 0 or -6g+6f+18+c=0(2)

and $\therefore {0}^{2} + {0}^{2} + 2 g \cdot 0 + 2 f \cdot 0 + c = 0$

$\therefore c = 0$ . Equation (1) is now $- 12 g + 36 + 0 = 0$

$\therefore 12 g = 36 \mathmr{and} g = 3$.Equation (2) is now

$- 6 g + 6 f + 18 + c = 0 \mathmr{and} - 6 \cdot 3 + 6 f + 18 + 0 = 0$ or

$- 18 + 6 f + 18 = 0 \mathmr{and} 6 f = 0 \therefore f = 0$ . Hence the equation of

circle is ${x}^{2} + {y}^{2} + 6 x = 0$

graph{x^2+y^2+6x=0 [-10, 10, -5, 5]}