Question

Write the equation of the circle containing the points J(-3, -3), K(1, -3), and L(1, 1)

Answer 1

general form of the circle,
$$x^2+y^2+cx+dy+e=0$$

since J,K,L go through the circle,
$$(-3)^2+(-3)^2+c(-3)+d(-3)+e=0\\
3c+3d-e=18\qquad(1)$$
$$(1)^2+(-3)^2+c(1)+d(-3)+e=0\\
3d-c-e=10\qquad(2)$$
$$(1)^2+(1)^2+c(1)+d(1)+e=0\\
c+d+e=-2\qquad(3)$$

3x(3)-(1)
$$4e=-24\implies e=-6$$
(2)+(3)
$$4d=8\implies d=2$$
$$c=2$$
equation of the circle-
$$x^2+y^2+2x+2y=6$$

Answer 2

thanks