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If $A=(1,2), B=(2,1)$ and $P$ is any point satisfying the condition $P A+P B=3$, then the equation of the locus of $P$ is
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The correct answer is:
$32 x^2+8 x y+32 y^2-108 x-108 y+99=0$
(c) Let the point $P(x, y)$ such that for two points $A(1,2)$ and $B(2,1)$,
$P A+P B=3$
$\Rightarrow \sqrt{(x-1)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-1)^2}=3$
$\Rightarrow \sqrt{(x-1)^2+(y-2)^2}=3-\sqrt{(x-2)^2+(y-1)^2}$
On squaring both sides, we get
$\Rightarrow \quad x^2-2 x+y^2-4 y+5=9+x^2-4 x+y^2$ $-2 y+5-6 \sqrt{(x-2)^2+(y-1)^2}$
$\Rightarrow \quad 2 x-2 y-9=-6 \sqrt{(x-2)^2+(y-1)^2}$
Again on squaring both sides, we get
$4 x^2+4 y^2+81-8 x y-36 x+36 y=36$ $\left[x^2+y^2-4 x-2 y+5\right]$
$\Rightarrow \quad 32 x^2+8 x y+32 y^2-108 x-108 y+99=0$
$P A+P B=3$
$\Rightarrow \sqrt{(x-1)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-1)^2}=3$
$\Rightarrow \sqrt{(x-1)^2+(y-2)^2}=3-\sqrt{(x-2)^2+(y-1)^2}$
On squaring both sides, we get
$\Rightarrow \quad x^2-2 x+y^2-4 y+5=9+x^2-4 x+y^2$ $-2 y+5-6 \sqrt{(x-2)^2+(y-1)^2}$
$\Rightarrow \quad 2 x-2 y-9=-6 \sqrt{(x-2)^2+(y-1)^2}$
Again on squaring both sides, we get
$4 x^2+4 y^2+81-8 x y-36 x+36 y=36$ $\left[x^2+y^2-4 x-2 y+5\right]$
$\Rightarrow \quad 32 x^2+8 x y+32 y^2-108 x-108 y+99=0$
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