Question

two equal circles touch each other externally at C and AB is a common tangent to the circles. then <ACB = ?

Answer 1

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

Answer 2

Answer:

∠ACB = α + β = 90°

Step-by-step explanation:

Let P be a point on AB such that, PC is at right angles to the Line Joining the centers of the circles.

Note that, PC is a common tangent to both circles.

This is because tangent is perpendicular to radius at point of contact for any circle.

Let ∠PAC= α and ∠PBC = β.

PA = PC [lengths of the tangents from an external point C]

In a triangle CAP, ∠PAC = ∠ACP = α

similarly PB = CP and ∠PCB = ∠CBP = β

now in the triangle ACB,

∠CAB + ∠CBA + ∠ACB = 180°   [sum of the interior angles in a triangle]

α + β + (α + β) = 180°   (Since ∠ACB = ∠ACP + ∠PCB = α + β.

2α + 2β = 180°

α + β = 90°

∴ ∠ACB = α + β = 90°