Two circles touch each other externally at P. AB is a common tangent to the circles touching them at A and B. The value of ∠APB is

Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.let ∠CAP = α and ∠CBP = β. CA = CP [lengths of the tangents from an external point C].In a triangle PAC, ∠CAP = ∠APC = α Similarly CB = CP and ∠CPB = ∠PBC = β Now in the triangle APB, ∠PAB + ∠PBA + ∠APB = 180o [sum of the interior angles in a triangle] α + β + (α + β) = 180o 2α + 2β = 180o α + β = 90o Therefore, ∠APB = α + β = 90o