Here we will prove that if two circles touch each other, the point of contact lies on the straight line joining their centres.
Case 1: When the two circles touch each other externally.
Given: Two circles with centres O and P touch each other externally at T.
To prove: T lies on the line OP.
Construction: Draw a common tangent XY through the point of contact T. Join T to O and P.
Proof:
Statement |
Reason |
1. ∠OTX = 90° |
1. Radius OT ⊥ tangent XY. |
2. ∠PTX = 90° |
2. Radius PT ⊥ tangent XY. |
3. ∠OTX + ∠PTX = 180° ⟹ ∠OTP = 180° ⟹ OTP is a straight line ⟹ T lies on OP. (Proved) |
3. Adding statement 1 and 2. |
Case 2: When the two circles touch each other internally at T.
To prove: T lies on OP produced.
Construction: Draw a common tangent XY through the point of contact T. Join T to O and P.
Proof:
Statement |
Reason |
1. ∠OTX = 90° |
1. Radius OT ⊥ tangent XY. |
2. ∠PTX = 90° |
2. Radius PT ⊥ tangent XY. |
3. OT and PT are both ⊥ to XY at the same point T. |
3. From statement 1 and 2. |
4. OT and PT lies on the same straight line ⟹ OTP is a straight line ⟹ T lies on OP. (Proved) |
4. Only one perpendicular can be drawn to a line through a point on it. |
Note: Let two circles with centres O and P touch each other at T. Let OT = r1 and PT = r2 and r1 > r2.
Let the distance between their centres = OP = d.
It is clear from the figures that
• When the circles touch externally, d = r_{1} + r_{2}.
• When the circles touch internally, d = r_{1} - r_{2}.
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