MHB Finding the Difference between Two Intersecting Equations

[Albert1]

Two diagrams of equations :
(1)$x^2+y^2=4+12x+6y$
and
(2)$x^2+y^2=k+4x+12y$
will intersect only when:$a\leq k \leq b$
find:$b-a$

 

[Greg]

Spoiler

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 $$-36\le k\le104,b-a=104-(-36)=140$$

 

[Albert1]

Spoiler

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 $$-36\le k\le104,b-a=104-(-36)=140$$

very good solution !