Two circles touch each other externally at C and AB is a common tangent to the circles. Then, ACB=

Asked by  | 5th Jan, 2013, 08:56: PM

Expert Answer:

Let A be on a circle with 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. So the triangle 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.


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

Answered by  | 6th Jan, 2013, 01:19: PM

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