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A point moves, so that the sum of squares of its distance from the points (1,2) and (-2,1) is always $6 .$ Then, its locus is
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The correct answer is:
a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
Let $P$ be any point, whose coordinate is $(h, k)$
Given.
P moves, so that the sum of squares of its distances from the points $A(1,2)$ and $B(-2,1)$ is $6 .$
1.e., $(P A)^{2}+(P B)^{2}=6$
$\Rightarrow \quad(h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6$
$\Rightarrow \quad h^{2}+1-2 h+k^{2}+4-4 k+h^{2}+4+4 h$
$$
+k^{2}+1-2 k=6
$$
$\Rightarrow \quad 2 h^{2}+2 k^{2}+2 h-6 k+4=0$
$\Rightarrow \quad h^{2}+k^{2}+h-3 k+2=0$
$\therefore$ Required locus is
$$
x^{2}+y^{2}+x-3 y+2=0
$$
Which represent a circle. Whose centre is $\left(\frac{-1}{2}, \frac{3}{2}\right)$ and radius $=\sqrt{\frac{1}{4}+\frac{9}{4}-2}=\sqrt{\frac{5}{2}-2}=\frac{1}{\sqrt{2}}$
Given.
P moves, so that the sum of squares of its distances from the points $A(1,2)$ and $B(-2,1)$ is $6 .$
1.e., $(P A)^{2}+(P B)^{2}=6$
$\Rightarrow \quad(h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6$
$\Rightarrow \quad h^{2}+1-2 h+k^{2}+4-4 k+h^{2}+4+4 h$
$$
+k^{2}+1-2 k=6
$$
$\Rightarrow \quad 2 h^{2}+2 k^{2}+2 h-6 k+4=0$
$\Rightarrow \quad h^{2}+k^{2}+h-3 k+2=0$
$\therefore$ Required locus is
$$
x^{2}+y^{2}+x-3 y+2=0
$$
Which represent a circle. Whose centre is $\left(\frac{-1}{2}, \frac{3}{2}\right)$ and radius $=\sqrt{\frac{1}{4}+\frac{9}{4}-2}=\sqrt{\frac{5}{2}-2}=\frac{1}{\sqrt{2}}$
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