Two circles touch externally at P. AB is the common tangent to the ciecles touching them at A and B.
The value of Angle APB is
(a) 30
(b) 45
(c) 60
(d) 90
Plz explain me the reason
given: C1 and C2 are two circles touch each other externally at P. AB is the common tangent to the circles C1 and C2 at point A and B respectively.
TPT:∠APB= 90 deg.
Proof: let ∠CAP=x and ∠CBP=y
CA=CP [lengths of the tangents from an external point C]
therefore in triangle PAC, ∠CAP=∠APC=x
similarly CB=CP and ∠CPB=∠PBC=y
now in the triangle APB,
∠PAB+∠PBA+∠APB=180 [sum of the interior angles in a triangle]
x+y+(x+y)=180
2x+2y=180
x+y=90
therefore ∠APB=x+y=90
hence AB subtends a right angle at P