ADDM Circles

Q1)Join the centers of the two circles with a straight line and extend it such that it intersects with the tangents at ( a , 0 ).
The centers of x2+y2-6x-16=0: ( 3 , 0 )
The radius: 5
The centers of x2+y2+6x-40=0: ( - 3 , 0 )
The radius: 7
Then you can find a pair of similar triangles by drawing the radius of the circles ┴ to the tangents.
So,
5 / 7 = ( a – 3 ) / ( a + 3 )
a = 18
Suppose the tangent is y = mx + c,
0 = 18m + c
c = -18m
Then y = mx – 18m
Consider any of the circles:
7 = l [ ( m )( - 3 ) + ( - 1 )( 0 ) – 18m ] / √( m2 + 1 ) l
49 ( m2 + 1 ) = ( - 21m )2
392m2 = 49
m = ±√2/4
So the tangents are y = √2x/4 - 9√2/2 and y = -√2x/4 + 9√2/2
Q2) Is P a variable pt. on AB? If so,
AB: ( C2 – C1 )
x2+y2-10x-4y+19-x2-y2+4x-2y-1=0
6x-6y+18=0
x-y+3=0
Centres of C1: ( 2 , - 1 ) , centers of C2: ( 5 , 2 )
So,
Distance from P to ( 2 , - 1 ):
l [(1)(2)+(-1)(-1)+3]/√(12+(-1)2)l
=6/√2
Distance from P to ( 5, 2 ),
l[(1)(5)+(-1)(2)+3]/√2l
=6/√2
(6/√2)/(6√2)=1/k
k=1