2 circles touch each other externally at C & AB is common tangent to the circles.Then angle ACB=?

Let A be on a circle woth centre O and B be the point on
the circle with O' as centre. And AB be the tangent to both circles touching at A and
B.

Let the two circles touch at
C.

Let the tangent at C meet AB at
N.

Now NA and NT are tangents to the the circle with centre
O andtherefore NA= NB. Sothetriangle NAC is isosceles and angles NAC = NCA = x
say.

By similar consideration NB and NT are tangents from N
to circle with centre O'. So triangle NBC is isosceles with NB=NC and therefore, angles
NBC = NCB = y say.

Therefore in triangle ABC, angles A+B+C
= x + y + (x+y) = 180

Or 2(x+y)
=180.

x+y = 180/2 =
90.

Therefore,

x+y = angle ACB
=180/2 =90 degree.