Examples

Chapter 10 Class 10 Circles
Serial order wise

### Transcript

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