Two circles C(O, r) and C^{1}(O^{1}, r^{1}) touch each other at P, externally or internally.
Construction: join OP and O^{1}P .
Proof : we know that if two circles touch
each other, the point of contact lies on the line through the centres.
O, P,O^{1} are collinear
OPO^{1} is a sttraight line.
Case I: C(o,r) and C^{1}(o^{1},r^{1} )touch externally at P.
In ΔOPQ, OP=OQ= r
∠OQP =∠OQP (Angle opposite to equal sides)
Similarly in ΔO^{1}PR, O^{|}P= O^{|}R=r^{|}
∠ORP = ∠RPO -------(2)
Also ∠OPQ = ∠RPO -------(3) (Vertically opposite angles)
From (1),(2) and(3) we get
∠OQP = ∠RPO
But these are alternate angles
=> OP u O1R
Case I I: C(o,r) and C(o^{1},r^{1}) touch internally at P .
In ΔOPQ, OP=OQ= r
∠OQR = ∠O^{|}RP ...................(4) (angles opposite to equal side)
Similirarly in ΔO^{|}PR, O^{|}P = O^{|}R =r^{|}
∠ORP = ∠O^{1}PR ...................(5)
From (4) and (5) we get ∠OQR = ∠O^{|}RP
But these are correspondingg angles.
OQ // O^{|}R