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

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Answer 1 · 2018-03-19T13:05:18

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

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

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

$:. (-6)^2+0^2+2g*(-6)+2f*0 +c= 0$

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

$:. (-3)^2+3^2+2g*(-3)+2f*3 +c= 0$

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

and $:. 0^2+0^2+2g*0+2f*0 +c= 0$

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

$:. 12g=36 or g=3$ .Equation (2) is now

$-6g+6f+18+c=0 or -6*3+6f+18+0=0$ or

$-18+6f+18=0 or 6f=0 :. f=0$ . Hence the equation of

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

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