In the figure AB is a tangent to a circle with centre O. AP is a chord of the circle such that AP=PQ. Prove that OA² =OB*OQ?

Question

In the figure AB is a tangent to a circle with centre O AP is a
chord of the circle such that AP=PQ Prove that OA2
​=OB*OQ?

Priyanka Kedia · Accepted Answer

Dear Student,

In △OAQ and △OQP,OA=OP  radius of the circleOQ=OQ commonAQ=PQ  Given⇒△OAQ≅△OQP⇒∠AQO=∠PQOBut ∠AQO+∠PQO=180⇒∠AQO=∠PQO=90Now,In right △OAB,OB2=OA2+AB2⇒OB2=OA2+BQ2+AQ2     In right △AQB, AB2=BQ2+AQ2⇒OB2=OA2+OB-OQ2+AO2-OQ2        In right △AQO, AQ2=AO2+OQ2⇒OB2=OA2+OB2+OQ2-2OB OQ+AO2-OQ2⇒OA2=2OB OQHence Proved

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In the figure AB is a tangent to a circle with centre O. AP is a chord of the circle such that AP=PQ. Prove that OA2 ​=OB*OQ?

In the figure AB is a tangent to a circle with centre O. AP is a chord of the circle such that AP=PQ. Prove that OA² =OB*OQ?