ce amaths question

(a)
Let Q(h,k) be the common point,
Equation of tangent for C1 at (h,k) is
(x-a1)(h-a1) + (y-b1)(k-b1) = r1^2
It passes through (0,0),
Therefore,
-a1(h-a1)-b1(k-b1) = (r1)^2
(a1)^2 + (b1)^2 -(a1)h -(b1)k =(r1)^2 ... (1)
Similarly,
(a2)^2 + (b2)^2 -(a2)h -(b2)k = (r2)^2 ... (2)
(1)-(2),
(a1)^2-(a2)^2 + (b1)^2-(b2)^2 + (a2-a1)h + (b2-b1)k = (r1)^2-(r2)^2
Now, OQ is perpendicular to AB,
Then k/h * (b2-b1)/(a2-a1)= -1
k(b2-b1) = -h(a2-a1)
(a2-a1)h + (b2-b1)k = 0
Therefore,
(a1)^2-(a2)^2+(b1)^2-(b2)^2=(r1)^2-(r2)^2

(b)
Suppose the two tangents passes through R, S on the circles C1 and C2
Then angle ROS = 90 deg
then angle AOB = 1/2 angle ROS = 45 deg
Let AO makes an angle AOX with x-axis and BO makes and angle BOX with the x-axis.
angle AOB = angle AOX - angle BOX
tan (angle AOB)
= [tan (angleAOX) - tan (angle BOX)]/[1 + tan (angle AOX) tan (anle BOX)]
= [b1/a1 - b2/a2] / [1 + (b1/a1)(b2/a2)]
= [a2b1-a1b2] / [a1a2 +b1b2]
Since tan (angle AOB) = tan (45deg) = 1
Therefore,
[a2b1-a1b2] / [a1a2 +b1b2] = 1
[a2(b1)-a1(b2)]=[(a1(a2)+b1(b2)]