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(B) Let $P(h, k)$ be any point on the locus.Given that the sum of the squares of its distances from $A(1, 2)$ and $B(-2, 1)$ is $6$.$(PA)^2 + (PB)^2 =

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a circle with centre $(-\frac{1}{2}, \frac{3}{2})$ and radius $\frac{1}{\sqrt{2}}$

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(B) Let $P(h, k)$ be any point on the locus.Given that the sum of the squares of its distances from $A(1, 2)$ and $B(-2, 1)$ is $6$.$(PA)^2 + (PB)^2 = 6$$(h - 1)^2 + (k - 2)^2 + (h + 2)^2 + (k - 1)^2 = 6$$(h^2 - 2h + 1) + (k^2 - 4k + 4) + (h^2 + 4h + 4) + (k^2 - 2k + 1) = 6$$2h^2 + 2k^2 + 2h - 6k + 10 = 6$$2h^2 + 2k^2 + 2h - 6k + 4 = 0$$h^2 + k^2 + h - 3k + 2 = 0$Replacing $(h, k)$ with $(x, y)$, the locus is $x^2 + y^2 + x - 3y + 2 = 0$.This is the equation of a circle.The centre is $(-\frac{1}{2}, \frac{3}{2})$ and the radius is $\sqrt{(-\frac{1}{2})^2 + (\frac{3}{2})^2 - 2} = \sqrt{\frac{1}{4} + \frac{9}{4} - 2} = \sqrt{\frac{10}{4} - 2} = \sqrt{\frac{5}{2} - 2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

$A$ point moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is always $6$. Then, its locus is

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