If A=(1,2), B=(2,1) and P is a variable point | vedclass

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(C) Let the coordinates of the variable point be $P(h, k)$.Given the coordinates $A(1, 2)$ and $B(2, 1)$,the condition is $|PA-PB|=3$.$\sqrt{(h-1)^2+(

Vedclass · Accepted Answer

$32x^2+8xy+32y^2-108x-108y+99=0$

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(C) Let the coordinates of the variable point be $P(h, k)$.Given the coordinates $A(1, 2)$ and $B(2, 1)$,the condition is $|PA-PB|=3$.$\sqrt{(h-1)^2+(k-2)^2} - \sqrt{(h-2)^2+(k-1)^2} = \pm 3$.Squaring both sides after isolating one radical:$\sqrt{(h-1)^2+(k-2)^2} = 3 + \sqrt{(h-2)^2+(k-1)^2}$.$(h-1)^2+(k-2)^2 = 9 + (h-2)^2+(k-1)^2 + 6\sqrt{(h-2)^2+(k-1)^2}$.$h^2-2h+1 + k^2-4k+4 = 9 + h^2-4h+4 + k^2-2k+1 + 6\sqrt{(h-2)^2+(k-1)^2}$.$2h - 2k - 9 = 6\sqrt{(h-2)^2+(k-1)^2}$.Squaring again:$(2h-2k-9)^2 = 36((h-2)^2+(k-1)^2)$.$4h^2+4k^2+81-8hk-36h+36k = 36(h^2-4h+4+k^2-2k+1)$.$4h^2+4k^2+81-8hk-36h+36k = 36h^2+36k^2-144h-72k+180$.$32h^2+32k^2+8hk-108h-108k+99=0$.Replacing $(h, k)$ with $(x, y)$,the locus is $32x^2+8xy+32y^2-108x-108y+99=0$.

If $A=(1,2), B=(2,1)$ and $P$ is a variable point satisfying the condition $|PA-PB|=3$,then the locus of $P$ is

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