What is the equation of the circle touching the lines: 2x +y-9=0, -2x +y -1=0 and -x +2y +7=0.
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.
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.
A plot of the three lines L1, L2 and L3 may be viewed here:
https://imagizer.imageshack.us/v2/500x300q90/674/pUzIJX.png
To find the equation of a circle that touches three given lines, you can use the concept of the circumcircle. The circumcircle of a triangle is a circle that passes through all three vertices of the triangle. In this case, the three lines represent the sides of the triangle, and the circle we want is its circumcircle.
To find the equation of the circumcircle, we need to find the center coordinates (h, k) and the radius r.
1. Start by finding the intersection points of the given lines. Solve each pair of equations to find the coordinates of the three intersection points.
Intersection of Line 1 and Line 2:
Solve the system of equations: 2x + y - 9 = 0 and -2x + y - 1 = 0
Intersection of Line 1 and Line 3:
Solve the system of equations: 2x + y - 9 = 0 and -x + 2y + 7 = 0
Intersection of Line 2 and Line 3:
Solve the system of equations: -2x + y - 1 = 0 and -x + 2y + 7 = 0
2. Once you have the coordinates of the three intersection points, use them to find the equations of the perpendicular bisectors of the sides of the triangle formed by those points.
For each side of the triangle, find the midpoint of the side by averaging the coordinates of the two endpoints. Then find the slope of the line passing through the two endpoints and take the negative reciprocal to get the slope of the perpendicular bisector. Use the midpoint and the slope to find the equation of the perpendicular bisector line.
3. Find the point of intersection of any two perpendicular bisectors. These two lines will intersect at the center of the circumcircle.
4. Once you have the center coordinates, you can find the radius. Calculate the distance between the center and any of the three intersection points. This distance is equal to the radius of the circle.
5. Finally, use the center coordinates (h, k) and the radius r to write the equation of the circle in the standard form: (x - h)^2 + (y - k)^2 = r^2.
By following these steps, you can find the equation of the circle that touches the given lines.