Locus

When a point's sum of squared distances to two fixed points is constant, expanding both squared distances causes the linear terms to cancel by symmetry, leaving a circle centred at the midpoint. Let the point be (x, y). Then (x - 3)^2 + y^2 + (x + 3)^2 + y^2 = 50. Expanding gives x^2 - 6x + 9 + y^2 + x^2 + 6x + 9 + y^2 = 50, so 2x^2 + 2y^2 + 18 = 50. Subtracting 18 yields 2x^2 + 2y^2 = 32, and dividing by 2 gives x^2 + y^2 = 16. Option x^2 + y^2 = 25 forgets to subtract the constant 18 properly. Option x^2 + y^2 = 9 mishandles the division by 2. Option x^2 + y^2 = 32 omits the final halving step. This is the standard JEE Advanced locus-as-circle derivation. Plausibility check: the locus is a circle of radius 4 centred at the origin, the midpoint of the two fixed points, exactly as symmetry predicts for such a sum-of-squares condition.