Question 800179
Looking at them from some point of view, the circles must look like this:
{{{drawing(300,150,-2.5,2.5,-1.25,1.25,
circle(-1,0,0.05),circle(1,0,0.05),
locate(-1.05,-0.06,A),locate(0.95,-0.06,B),
circle(-1,0,1),circle(1,0,1),
circle(0,0,0.05),locate(0.1,0.1,D)
)}}}
Point D is at the same distance from the center of both circles, and on the line that connects those centers.
It is the midpoint of segment AB.
Its x-coordinate is the average of the x-coordinates of A and B
The x-coordinate of B(8,7) is {{{x[B]=8}}}
We know that the x-coordinate of A is zero, because {{{x=0}}} for all points on the y-axis and A is on the y-axis. So {{{x[A]=0}}}
So, the x-coordinate of D(p,p) is {{{x[D]=(0+8)/2}}},
meaning that {{{highlight(p=4)}}}.
Since it is written as D(p,p), with the same {{{p}}} as x and y-coordinate, it must be D(4,4).
For A(0,{{{y[A]}}}), that would mean that {{{y[D]=4=(y[A]+y[B])/2=(y[A]+7)/2}}},
and from {{{(y[A]+7)/2=4}}} we get
{{{y[A]+7=4*2}}} --> {{{y[A]+7=8}}} --> {{{y[A]=8-7}}} --> {{{y[A]=1}}}
Also, the radius of the circles is the distance from B(8,7) to D(4,4).
That is
{{{sqrt((8-4)^2+(7-4)^2)=sqrt(4^2+3^2)=sqrt(16+9)=sqrt(25)=5}}}
Now, I could draw those circles, with the x- and y-axes, the line connecting their centers, and the common tangent to both circles.
{{{drawing(300,300,-6,14,-6,14,
grid(1),
blue(circle(0,1,5)),blue(circle(8,7,5)),
blue(circle(0,1,0.2)),blue(circle(8,7,0.2)),
blue(line(0,1,8,7)),green(line(-5,16,13,-8))
)}}}
The slope of AB is
{{{(7-1)/(8-0)=6/8=3/4}}}
The tangent to both circles is perpendicular to AB, so its slope is
{{{(-1)/((3/4))=-4/3}}}.
The tangent passes through D(4,4) so its equation is 
{{{y-4=(-4/3)(x-4)}}} --> {{{y-4=(-4/3)x+16/3}}} --> {{{y=(-4/3)x+16/3+4}}} --> {{{y=(-4/3)x+16/3+12/3}}} --> {{{highlight(y=(-4/3)x+28/3)}}}
We could also write it differently:
{{{y=(-4/3)x+28/3}}} --> {{{3y=-4x+28)}}} --> {{{highlight(4x+3y=28)}}}