Let the two given points be \(A=(a,0)\) and \(B=(-a,0)\). Let P be \((x,y)\).
Distance PA = \(\sqrt{(x-a)^2 + (y-0)^2} = \sqrt{(x-a)^2 + y^2}\).
Square of distance PA: \(PA^2 = (x-a)^2 + y^2 = x^2 - 2ax + a^2 + y^2\).
Distance PB = \(\sqrt{(x-(-a))^2 + (y-0)^2} = \sqrt{(x+a)^2 + y^2}\).
Square of distance PB: \(PB^2 = (x+a)^2 + y^2 = x^2 + 2ax + a^2 + y^2\).
Given that the sum of squares of distances is \(2b^2\):
\(PA^2 + PB^2 = 2b^2\).
\((x^2 - 2ax + a^2 + y^2) + (x^2 + 2ax + a^2 + y^2) = 2b^2\).
Combine terms:
\(2x^2 + 2y^2 + 2a^2 = 2b^2\).
Divide by 2:
\(x^2 + y^2 + a^2 = b^2\).
Rearrange to get the locus of P:
\(x^2 + y^2 = b^2 - a^2\).
This represents a circle centered at the origin \((0,0)\) with radius \(\sqrt{b^2-a^2}\), provided \(b^2>a^2\). If \(b^2=a^2\), it's a point (origin). If \(b^2<a^2\), no real locus.
This matches option (b).
\[ \boxed{x^2+y^2 = b^2-a^2} \]