Example 1(Method 1)
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Given: Let two concentric circles be C1 & C2
with center O
AB be chord of the larger circle C1
which touches the smaller circle C2 at point P
To prove: Chord AB is bisected at point of contact (P)
i.e AP = BP
Solution:
Since AB is tangent to smaller circle C2
OP ⊥ AB
Now ,
AB is a chord of bigger circle C1 and OP ⊥ AB
As perpendicular from the centre bisects the chord
∴ OP would be bisector of chord AB,
⇒ AP = BP
Hence proved
Example 1(Method 2)
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Given: Let two concentric circles be C1 & C2
with center O
AB be chord of the larger circle C1
which touches the smaller circle C2 at point P
To prove: Chord AB is bisected at point of contact (P)
i.e AP = BP
Solution:
Since AB is tangent to smaller circle C2
OP ⊥ AB
Since OP⊥ AB
∠ OPA = ∠ OPB = 90°
Hence Δ OPA and Δ OPB are right triangles
Join OA and OB
Using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
Example 1(Method 3)
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Given: Let two concentric circles be C1 & C2
with center O
AB be chord of the larger circle C1
which touches the smaller circle C2 at point P
To prove: Chord AB is bisected at point of contact (P)
i.e AP = BP
Solution:
Since AB is tangent to smaller circle C2
OP ⊥ AB
Join OA and OB
In Δ OAP and OBP
∠ OPA = ∠ OPB
OA = OB
OP = OP
∴ Δ OAP ≅ OBP
Hence, AP = BP
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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