The three lines intersect to form a triangle. Naturally, there is one circle inside the triangle, and three circles outside the triangle, all of which touch the three lines.
The three lines intersect at (2,5),(5,-1),(-3,-5). If we call those vertices A,B,C, then the opposite sides are a,b,c, and we have
Now, the incenter lies at the intersection of the angle bisectors of the vertices. The three lines have slopes -2,2,1/2. Thus, the angle bisectors at A,B,C have slopes undef,-1/3,1
So, the center lies on the line x = 2
The other lines are
y+1 = -1/3 (x-5)
y+5 = (x+3)
They intersect at (2,0)
The distance from (2,0) to the three original lines is √5, so the circle is
(x-2)^2 + y^2 = 5
See the plots at
http://www.wolframalpha.com/input/?i=plot+2x+%2By-9%3D0%2C+-2x+%2By+-1%3D0+%2C+-x+%2B2y+%2B7%3D0%2C+%28x-2%29^2+%2B+y^2+%3D+5
You can work similar magic if you want to find the excircles.
What is the equation of the circle touching the lines: 2x +y-9=0, -2x +y -1=0 and -x +2y +7=0.
3 answers
This problem is very interesting, and as Steve pointed out, there are 4 circles each of which is tangent to all three lines. This prompted me to look for a general solution for all four circles.
If we first examine the condition of tangency of a circle
C: (x-a)²+(y-b)²=r²
to the line
y=mx+q,
it turns out to be
(b-ma-q)²=r²(1+m²)
For the circle to be tangent to all three lines,
L1: y=m1x+q1
L2: y=m2x+q2
L3: y=m3x+q3
where
m1=-2 q1=9;
m2=2 q2=1;
m3=1/2 q3=-7/2;
We can set up the system of equation of three unknowns in a, b and r:
(b+2a-9)²=5r²
(b-2a-1)²=5r²
(b-a/2+7/2)²=5r²/4
The solution of which will give the various values of a,b and r.
In particular, we can take square-root on both sides to give:
b+2a-9=sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=sqrt(5)r/2
The solution of which is
a=2, b=-10, r=3sqrt(5) for the circle below all three lines L1, L2 and L3.
The set
b+2a-9=sqrt(5)r
b-2a-1=-sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=-13, b=5, r=6sqrt(5) for the circle to the left and above L2 & L3.
The set
b+2a-9=-sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=7, b=5, and r=2sqrt(5) for the circle to the right, and above L1 & L3.
Finally, the set
b+2a-9=sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=2, b=0 and r=sqrt(5) for the in-circle above L3, as obtained by Steve in the previous post.
If we first examine the condition of tangency of a circle
C: (x-a)²+(y-b)²=r²
to the line
y=mx+q,
it turns out to be
(b-ma-q)²=r²(1+m²)
For the circle to be tangent to all three lines,
L1: y=m1x+q1
L2: y=m2x+q2
L3: y=m3x+q3
where
m1=-2 q1=9;
m2=2 q2=1;
m3=1/2 q3=-7/2;
We can set up the system of equation of three unknowns in a, b and r:
(b+2a-9)²=5r²
(b-2a-1)²=5r²
(b-a/2+7/2)²=5r²/4
The solution of which will give the various values of a,b and r.
In particular, we can take square-root on both sides to give:
b+2a-9=sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=sqrt(5)r/2
The solution of which is
a=2, b=-10, r=3sqrt(5) for the circle below all three lines L1, L2 and L3.
The set
b+2a-9=sqrt(5)r
b-2a-1=-sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=-13, b=5, r=6sqrt(5) for the circle to the left and above L2 & L3.
The set
b+2a-9=-sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=7, b=5, and r=2sqrt(5) for the circle to the right, and above L1 & L3.
Finally, the set
b+2a-9=sqrt(5)r
b-2a-1=sqrt(5)r
b-a/2+7/2=-sqrt(5)r/2
gives a=2, b=0 and r=sqrt(5) for the in-circle above L3, as obtained by Steve in the previous post.
A plot of the three lines L1, L2 and L3 may be viewed here:
https://imagizer.imageshack.us/v2/500x300q90/674/pUzIJX.png
https://imagizer.imageshack.us/v2/500x300q90/674/pUzIJX.png
1 answer