Two circles touch each other externally at point C. A direct common tangent AB touches the two circles at points A and B. Find the value of ∠ACB.
Given X and Y are two circles touch each other externally at C. AB is the common tangent to the circles X and Y at point A and B respectively.
To find: ∠ACB
Proof: 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 the tangent is perpendicular to the radius at the 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°
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