If A(2, 3) and B(3, -2) are two fixed points | vedclass

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(C) Given the condition $|PA - PB| = 2$.The distance formula gives $PA = \sqrt{(x - 2)^2 + (y - 3)^2}$ and $PB = \sqrt{(x - 3)^2 + (y + 2)^2}$.So,$|\s

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$(x - 5y - 2)^2 = 4[(x - 3)^2 + (y + 2)^2]$

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(C) Given the condition $|PA - PB| = 2$.The distance formula gives $PA = \sqrt{(x - 2)^2 + (y - 3)^2}$ and $PB = \sqrt{(x - 3)^2 + (y + 2)^2}$.So,$|\sqrt{(x - 2)^2 + (y - 3)^2} - \sqrt{(x - 3)^2 + (y + 2)^2}| = 2$.This implies $\sqrt{(x - 2)^2 + (y - 3)^2} = 2 + \sqrt{(x - 3)^2 + (y + 2)^2}$.Squaring both sides:$(x - 2)^2 + (y - 3)^2 = 4 + (x - 3)^2 + (y + 2)^2 + 4\sqrt{(x - 3)^2 + (y + 2)^2}$.Expanding the squares:$x^2 - 4x + 4 + y^2 - 6y + 9 = 4 + x^2 - 6x + 9 + y^2 + 4y + 4 + 4\sqrt{(x - 3)^2 + (y + 2)^2}$.Simplifying the equation:$x^2 + y^2 - 4x - 6y + 13 = x^2 + y^2 - 6x + 4y + 17 + 4\sqrt{(x - 3)^2 + (y + 2)^2}$.$2x - 10y - 4 = 4\sqrt{(x - 3)^2 + (y + 2)^2}$.Dividing by $2$:$x - 5y - 2 = 2\sqrt{(x - 3)^2 + (y + 2)^2}$.Squaring again:$(x - 5y - 2)^2 = 4[(x - 3)^2 + (y + 2)^2]$.

If $A(2, 3)$ and $B(3, -2)$ are two fixed points and $P(x, y)$ is a variable point satisfying the condition $|PA - PB| = 2$,then the locus of $P$ is

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