The discussion focuses on the intersection of two equations: (1) $x^2+y^2=4+12x+6y$ and (2) $x^2+y^2=k+4x+12y$. The key finding is that these equations intersect when the parameter $k$ satisfies the condition $a \leq k \leq b$. The solution involves determining the values of $b$ and $a$, ultimately leading to the calculation of $b-a$. This provides a clear method for identifying the range of $k$ for which the equations intersect.
PREREQUISITESMathematicians, educators, students studying algebra and geometry, and anyone interested in the analysis of intersecting equations.
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 $$-36\le k\le104,b-a=104-(-36)=140$$