Question

Prove that the length of the tangents drawn from an external point to a circle are equal in measure?

Answer 1

Answer:

If the tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80∘, then ∠POA is equal to.

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Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

➜ A circle with centre O

➜ Two tangents PA and PB

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

➜ PA = PB

$\Large{\underline{\underline{\bf{Solution:}}}}$

➜ Join OA, OB , OP

➜ Consider ΔOAP and ΔOBP

    OP = OP(common)

   ∠OBP = ∠OAP (∵ OB⊥ BP, OP⊥ AP)

   OA = OB (radii)

➜ Hence Δ OAP ≅ ΔOBP (By RHS criteria)

➜ ∴ PA = PB (by CPCT)

Hence proved.

$\Large{\underline{\underline{\bf{More\:information:}}}}$

➜ The tangent at any point of a circle is perpendicular to the radius of circle through the point of contact.

➜The angle made by the chord and tangent is equal to the angle made in the alternate segments.

➜ The tangents drawn at the ends of a diameter of a circle are parallel.\