The locus of the middle points of the chords of the circle x 2 + y 2 = a 2 which subtend a right angle at the centre, is -
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Let (h, k) be the mid-point of a chord AB of the circle x 2 + y 2 = a 2 . Then, the equation of AB is
hx + ky – a 2 = h 2 + k 2 – a 2 [Using T = S ′ ]
⇒ hx + ky = h 2 + k 2 … (1)
The combined equation of OA and OB is
x 2 + y 2 = a 2 
⇒ (h 2 + k 2 ) 2 (x 2 + y 2 ) – a 2 (hx + ky) 2 = 0
OA and OB will be perpendicular if
Coefficient of x 2 + Coefficient of y 2 = 0
⇒ (h 2 + k 2 ) 2 – a 2 h 2 + (h 2 + k 2 ) 2 – a 2 k 2 = 0
⇒ 2(h 2 + k 2 ) 2 – a 2 (h 2 + k 2 ) = 0
⇒ 2(h 2 + k 2 ) – a 2 = 0.
So, locus of (h, k) is 2(x 2 + y 2 ) – a 2 = 0.
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