SOLUTION: find the equation of the circle inscribed in a triangle whose sides are 3x+y-5=0, x+3y-1=0, x-3y+7=0.

Question 707548: find the equation of the circle inscribed in a triangle whose sides are 3x+y-5=0, x+3y-1=0, x-3y+7=0.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The radius of the circle is the distance, , from center (h,k) to each of the lines.
The distance from a point (h,k) to a line can be calculated as
, so
which simplifies to
and
With those absolute values, there are too many possibilities, so we should figure out the sign of the expressions inside the absolute values.
It's easy to sketch the triangle and figure out.
With the line equations in that form, it's easy to calculate intercepts:
For example, wit ,
--> --> and
--> --> -->
So, goes through (5/3,0) and (0,5).
similarly,
goes through (1,0) and (0,1/3),
and goes through (-7,0) and (0,7/3).
Sketching, we can find what points are inside the triangle,
including point (h,k), the center of the circle.
The triangle is the space below the lines that go through (0,5) and (-7,0)
(the first and third lines), but above the line for (the with intercepts so close to the origin.
Each of the expressions inside the absolute values would be zero if (h,k) is on the corresponding line, positive if (h,k) is on one side of the line, and negative if it is on the other side. Since none of the lines goes through the origin, we can use the origin as a test point to figure out which side is which.

is true for the part of the plane below ,
where we find the origin, (0,0), the triangle, and (inside the triangle)
the center of the circle (h,k).
So, and

is true for the origin and the part of the plane below ,
but for the other side, where the triangle and (h,k) are.
So, and
is true for the origin and the part of the plane below ,
where we find the origin, (0,0), the triangle, and point (h,k).
So, and

Now that we know the sign of each expression, we can re-write
as

and that is a system of equations that we can solve to get
, and
--> -->

Then we can write the equation of the circle as

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