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 Prove that `4q^(2)=p^(2)+3r^(2)`.

Explanation

Let `AC =x`. Then, `AB = 2x`.
Let `OL _|_ AB ` and `OM _|_ AC`
The, `OL =p, OM =q` and `OA = r`
Also, `AM = (1)/(2) AC = (x) /(2)` and `AL = (1)/(2) AB =x`
Clearly, `r^(2) = p^(2) x^(2)` ...(i) ltbr and `r^(2) = q^(2) + ((x)/(2))^(2) ` .......(ii)
`:. q^(2) = r^(2)- (x^(2))/(4) implies 4q^(2) = 4r^(2) -x^(2) `
`implies 4q^(2) = 4r^(2) - (r^(2) -p^(2) ) ` [ using (i) ]
`=4q^(2) = 3r^(2) +p^(2)`
Hence, `4q^(2) = 3r^(2) +p^(2)`