For the complete understanding of the solution of this question please check the two attached files too.
Let us begin.
There are two circles given to us. Let us call them [tex] C1 [/tex] and [tex] C2 [/tex]. Now, we know that [tex] C1 [/tex] has centre at [tex] (2,1) [/tex] and has a radius, say, [tex] r_{1} [/tex] of 1. We also know that [tex] C2 [/tex] has centre at [tex] (8,9) [/tex] and has a radius, say, [tex] r_{2} [/tex] of 9.
We know from coordinate geometry that of the center [tex] (a,b) [/tex] and the radius, [tex] r [/tex] of a circle are given then, the equation of the circle is given by:
[tex] (x-a)^2+(y-b)^=r^2 [/tex]
Using this formula and applying it to both the circles, we will get the equations of the two circles to be:
[tex] C1:(x-2)^2+(y-1)^2=1^2 [/tex]
or [tex] (x-2)^2+(y-1)^2-1=0 [/tex] (subtracting both sides by 1)
and
[tex] C2: (x-8)^2+(y-9)^2=9^2 [/tex]
or [tex] (x-8)^2+(y-9)^2-81=0 [/tex] (subtracting both sides by [tex] 9^2 [/tex] which is 81).
Now, we know that, at if the circles have a common external tangent such that [tex] m<0 [/tex] or in other words [tex] m [/tex] is negative then we must be having the condition that the circles touch each other as is evident from the first diagram.
All we need to do now is to equate the two equations of the circles [tex] C1 [/tex] and [tex] C2 [/tex] and solve for [tex] y [/tex] in terms of [tex] x [/tex].
That will give us:
[tex] (x-2)^2+(y-1)^2-1=(x-8)^2+(y-9)^2-81=0 [/tex]
Solving this we get:
[tex] 4y=-3x+15 [/tex]
[tex] y=-\frac{3}{4}x +\frac{15}{4} [/tex]
[tex] y=-0.75x+3.75 [/tex]
This is the required equation of the tangent line and thus, as we can see b=[tex] \frac{15}{4} [/tex] or [tex] 3.75 [/tex]
