very good solution !greg1313 said:We have $$x^2+y^2=4+12x+6y\Rightarrow(x-6)^2+(y-3)^2=49$$ and $$x^2+y^2=k+4x+12y\Rightarrow(x-2)^2+(y-6)^2=k+40$$The distance between these two circles is 5 units, with one of the circles having a radius of 7. This means the other circle must have a radius of at least 2, giving a lower bound on k of $-36$. Since this other circle remains in contact with the circle of radius 7 until its radius is 12, an upper bound on k is 104. So we have $$-36\le k\le104,b-a=104-(-36)=140$$
The difference between two intersecting equations is the value or set of values that satisfy one equation but not the other. In other words, it is the solution to one equation that is not a solution to the other.
To find the difference between two intersecting equations, you can set the equations equal to each other and solve for the variable. The resulting value(s) will be the difference between the two equations.
If there is no difference between two intersecting equations, it means that the equations have the same solution(s). This could also mean that the equations are equivalent or represent the same line or curve.
Yes, it is possible for two intersecting equations to have more than one difference. This can happen if the equations are not linear or if they have multiple solutions that satisfy one equation but not the other.
Finding the difference between two intersecting equations can be useful in various fields such as mathematics, physics, and engineering. It can help determine the unique solutions to a system of equations and can be used to solve real-world problems involving intersecting lines or curves.