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Question

# AB and AC are two chords of a circle of radius r such that AB=2AC. If p and q are the distances of AB and AC from the centre, then show that 4q2=p2+3r2.

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Solution

## Consider the diagram shown below. Given: AB and AC are two chords of a circle with center O. Such that AB=2AC p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q r is the radius of the circle To prove that: 4q2=p2+3r2 Proof: Join OA. OM and ON are ⊥ distances of AB and AC from center O. Here, AN=AC2 (perpendicular from center to chord intersect at mid-point of the chord) AM=AB2 (perpendicular from center to chord intersect at mid-point of the chord) In right angled ΔOMA, OM2+AM2=OA2 p2+AM2=r2 AM2=r2−p2 …… (1) In right angled ΔONA, ON2+AN2=OA2 q2+AN2=r2 AN2=r2−q2 …… (2) Since, AM=AB2=2AC2=AC=2AN From equations (1) and (2), we have r2−p2=AM2 r2−p2=4AN2 r2−p2=4[r2−q2] 4q2=p2+3r2 LHS=RHS Hence, proved.

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