anothr cet q

No man thats a bad idea.
I know a quicker one.
We know the condition for tangency,

c=±√a²m²+b²
Substituting in y=mx+c,
lets rearrange the equation to get a quadratic in m.
m²(x²-a²)-2xym+y²+b²=0
From the theory of quadratic equations we have,
Let roots be m₁ and m₂.
Now m₁+m₂=2xy/(x²-a²)
Now put m₁+m₂=5,
And we get,
5x²-5a²-y²-b²=0

Voila !!
You locus is ready.
:)

nice one :)
only check your 'c'..