Let us choose the fixed point P as the origin. Let us now choose any line through P making an angle θ with + ve direction of the X- axis (see fig).

Any point on this line can be chosen as

(r cos θ , r sin θ ) where r is the distance measured from P.

If this point must also lie on the given curve, then we have

r 2 (a cos 2 θ + h sin 2 θ + h sin 2 θ + b sin 2 θ ) – 1 = 0 ... (i)

For any given θ , therefore there will be two values of r.

If PQ and PR be the roots of equation (i), then we have

PQ. PR = – where f( θ ) = a cos 2 θ + h sin 2 θ + b sin 2 θ

Since f( θ ) is given to be independent of θ , therefore using calculus, we have

= – 2 a sin θ cos θ + 2h cos 2 θ + 2b sin θ cos θ = 0

i.e. (b – a) sin 2 θ + 2h cos 2 θ = 0 ... (ii)

Identically. (The word identically here implies, true for every value of θ ).

Equation (ii) will be identically true if and only if a = b and h = 0

which therefore proves that the given curve must be circle.