Consider lines $ax+2by+b=2a+3by$ a, b$\in$R, then?
- A unique line of above family is tangent to circle $x^2+y^2+x-y-6=0$
- Two lines from above family are tangent to circle $x^2+y^2-6x-4y-1=0$
- If above lines are pair of tangents to $x^2+y^2-x-y-1=0$, then length of tangent is $1$ unit
- Circumcircle of triangle formed by above lines are pair of tangents of $x^2+y^2+4x+2y+1=0$ and corresponding chord of contact is $x^2+y^2=5$
Rearranging the given line equation ax + 2by + b = 2a + 3by yields (x - 2)a + (3b - 2)y + b = 0, which represents a family of lines passing through the intersection of x - 2 = 0 and 3b - 2 = 0, or alternatively can be parameterized by substituting a/b to find the envelope. For the circle x squared plus y squared plus x minus y minus 6 equals 0, its center is (-0.5, 0.5) and its radius is the square root of 6.5, which is approximately 2.55. Testing the family of lines, which all pass through the fixed point (2, -2), shows that the distance from the center (-0.5, 0.5) to the point (2, -2) is exactly the square root of 6.5, meaning the point of intersection lies exactly on the circle. Therefore, exactly one unique line from this family acts as a tangent to the circle.