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Question

# Question 12 AB and AC are two chords of a circle of radius r such that AB = 2AC. If p an q are the distances of AB an AC from the centre, prove that 4q2=p2+3r2.

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Solution

## Given in a circle of radius r, there are two chords AB and AC such that AB = 2AC. Also, the distance of AB and AC from the centre are p and q, respectively. To prove that 4q2=p2+3r2 Proof Let AC = a, then AB = 2a From centre O, perpendicular is drawn to the chords AC and AB at M and N, respectively ∴ AM=MC=a2 AN = NB = a In Δ OAM AO2=AM2+MO2 [by Pythagoras theorem] AO2=(a2)2+q2 .........(i) In Δ OAN, AO2=(AN)2+(NO)2 [by Pythagoras theorem] AO2=a2+p2 .............(ii) From Eqs. (i) and (ii) (a2)2+q2=a2+p2 ⇒ a24+q2=a2+p2 ⇒ a2+4q2=4a2+4p2 [multiplying both sides by 4] ⇒ 4q2=3a2+4p2 ⇒ 4q2=p2+3(a2+p2) ⇒ 4q2=p2+3r2 [in right angled Δ OAN, r2=a2+p2] Hence proved.

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