SOLUTION: Find the equation of the circle that is inscribed in the triangle whose sides lie on the lines 3x+y=5,x-3y=1,and x+3y=-7.

Given three lines, there are 4 circles which are tangent to all
three lines.  There are three excircles and one inscribed circle
inside the triangle.  Here they are:



You only want the little red circle in the triangle, not the
3 excircles.  However, we can't tell which of the four is the
inscribed circle.  So we have to find them all.

Let the center of the circle be (h,k), and the radius r.  Then the
equation of the circle is (x-h)² + (y-k)² = r²

We use the formula for perpendicular distance d 

from the point whose coordinates are (x1,y1)

to the line whose equation is 

Ax + By + C = 0 

That is given by this formula:

d = 

The perpendicular distances from the center to the three lines
are all equal, since they are all three radii.

So we find all three distances from the center (h,k) to the 
three lines and we have:

We get 0 on the right side of the equations of the 
three lines

3x+y=5, x-3y=1, and x+3y=-7  

3x+y-5=0, x-3y-1=0, x+3y+7=0

Substituting in the formula above

r =  =  = 

r =  =  = 

So we have this system of equations:

 = 
 = 

We multiply them both by √10 

|3h + k - 5| = |h - 3k - 1|
|h - 3k - 1| = |h + 3k + 7|

There are four solutions to this system.  Here are the four systems

, , , 

Those simplify to

, ,  ,

Their solutions are

(,-4/3), (-3,), (, ), (-3,-9)

To find the radii, we substitute each of those in:

r = 

I'll just find the radius of the small red circle since that's
the only one you were asked to find.

It is the circle with the center (, ).
 
r = 

r = 

Multiply top and bottom by 6

r = 

r = 

r = 

r² = 

r² = 

Substituting in 

(x-h)² + (y-k)² = r²

(x-)² + (y-())² = 

(x-)² + (y+)² = 

Or you can multiply that out and simplify it in general form
and get:

360x² + 360y² - 600x + 960y + 361 = 0

It doesn't come out very nice but it's correct.

Edwin

Given three lines, there are 4 circles which are tangent to all
three lines.  There are three excircles and one inscribed circle
inside the triangle.  Here they are:



You only want the little red circle in the triangle, not the
3 excircles.  However, we can't tell which of the four is the
inscribed circle.  So we have to find them all.

Let the center of the circle be (h,k), and the radius r.  Then the
equation of the circle is (x-h)² + (y-k)² = r²

We use the formula for perpendicular distance d 

from the point whose coordinates are (x1,y1)

to the line whose equation is 

Ax + By + C = 0 

That is given by this formula:

d = 

The perpendicular distances from the center to the three lines
are all equal, since they are all three radii.

So we find all three distances from the center (h,k) to the 
three lines and we have:

We get 0 on the right side of the equations of the 
three lines

3x+y=5, x-3y=1, and x+3y=-7  

3x+y-5=0, x-3y-1=0, x+3y+7=0

Substituting in the formula above

r =  =  = 

r =  =  = 

So we have this system of equations:

 = 
 = 

We multiply them both by √10 

|3h + k - 5| = |h - 3k - 1|
|h - 3k - 1| = |h + 3k + 7|

There are four solutions to this system.  Here are the four systems

, , , 

Those simplify to

, ,  ,

Their solutions are

(,-4/3), (-3,), (, ), (-3,-9)

To find the radii, we substitute each of those in:

r = 

I'll just find the radius of the small red circle since that's
the only one you were asked to find.

It is the circle with the center (, ).
 
r = 

r = 

Multiply top and bottom by 6

r = 

r = 

r = 

r² = 

r² = 

Substituting in 

(x-h)² + (y-k)² = r²

(x-)² + (y-())² = 

(x-)² + (y+)² = 

Or you can multiply that out and simplify it in general form
and get:

360x² + 360y² - 600x + 960y + 361 = 0

It doesn't come out very nice but it's correct.

Edwin