When it is said a circle circumscribed about a triangle, is it mean the circle lie in the triangle or vice versa? I see the text in a question: 
Find the equation of the circle circumscribed about the triangle formed by the coordinate axes and the line $x+2y=2$!
From the problem, I have two thinkings:
When it is the triangle lies in the circle, it is easy to solve the problem since we can find the desired equation by substituting each of the three intersectional points (the vertices of the triangle) into a circle equation $x^2 + y^2 + Cx + Dy + E = 0$ and solve the equation system.
However, what about if the circle lies in the triangle?
 A: The circumcircle lies outside of the triangle - is circumscribed around it - passing through the three vertices.  Its center is located at the intersection of the perpendicular bisectors of the edges.
The incircle lies inside of the triangle - is inscribed within it - tangent to all three edges.  Its center is located at the intersection of the angle bisectors of the three vertices.

I guess that answered only the title of the question.  Let's get to the true question: how do you get the incircle?
For this the easiest path I think is to get the incenter, the center of the incircle.  With $a$ as the length of the segment opposite of vertex $A$, etc, the incenter is located at
$$\frac{aA + bB + cC}{a + b + c}$$
Which is one of the most elegant formulas I've seen all year.
Then, the inradius is equal to
$$\left|\frac{A \times B + B \times C + C \times A}{a + b + c}\right|$$
Which is almost as good as that last one.  (the $\times$ symbol here represents the scalar analogue to cross product: $A \times B = A_xB_y - A_yB_x$)
A: Combine K and $L_1$ to form one quadratic.
Repeat for $L_2$ and $L_3$.
Then we have 3 equations for 3 unknowns.
Remark:- The general equation of a circle, $K : x^2 + y^2 + Dx + Ey + F = 0$ has three unknowns only.
A: There is another way to get the equation for the circumcircle for this problem:
The coordinate axes intersect the line $x+2y=2$ at $(0,1)$ and $(2,0)$.  Since the coordinate axes are perpendicular at $(0,0)$, the line segment connecting $(0,1)$ and $(2,0)$ is a diameter of the circumcircle, so the center is at $(1,{1\over2})$, and thus the equation of the circumcircle is
$$(x-1)^2+\left(y-{1\over2}\right)^2=1^2+\left(1\over2\right)^2$$
which simplifies to
$$x^2+y^2=2x+y$$
