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If $A(2,3)$ and $B(3,-2)$ are two fixed points and $P(x, y)$ is a variable point satisfying the condition $|P A-P B|=2$, then the locus of $P$ is
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Verified Answer
The correct answer is:
$(x-5 y-2)^2=4\left[(x-3)^2+(y+2)^2\right]$
According to question,
$\begin{gathered}
|P A-P B|=2 \\
\Rightarrow \sqrt{(x-2)^2+(y-3)^2}-\sqrt{(x-3)^2+(y+2)^2}=2 \\
\Rightarrow \sqrt{(x-2)^2+(y-3)^2}=2+\sqrt{(x-3)^2+(y+2)^2}
\end{gathered}$
On squaring, we get
$\begin{aligned}
& \begin{array}{c}
(x-2)^2+(y-3)^2=4+(x-3)^2 \\
+(y+2)^2
\end{array} \\
& \begin{aligned}
\Rightarrow \quad & +4 \sqrt{(x-3)^2+(y+2)^2} \\
& x^2+4-4 x+y^2+9-6 y
\end{aligned} \\
& \quad 4+x^2+9-6 x+y^2+4+4 y+4 \\
& \Rightarrow \quad 2 x-10 y-4=4 \sqrt{(x-3)^2+(y+2)^2} \\
& \Rightarrow \quad x-5 y-2=2 \sqrt{(x-3)^2+(y+2)^2} \\
& \text { On squaring, }(x-5 y-2)^2=4\left[(x-3)^2+(y+2)^2\right]
\end{aligned}$
$\begin{gathered}
|P A-P B|=2 \\
\Rightarrow \sqrt{(x-2)^2+(y-3)^2}-\sqrt{(x-3)^2+(y+2)^2}=2 \\
\Rightarrow \sqrt{(x-2)^2+(y-3)^2}=2+\sqrt{(x-3)^2+(y+2)^2}
\end{gathered}$
On squaring, we get
$\begin{aligned}
& \begin{array}{c}
(x-2)^2+(y-3)^2=4+(x-3)^2 \\
+(y+2)^2
\end{array} \\
& \begin{aligned}
\Rightarrow \quad & +4 \sqrt{(x-3)^2+(y+2)^2} \\
& x^2+4-4 x+y^2+9-6 y
\end{aligned} \\
& \quad 4+x^2+9-6 x+y^2+4+4 y+4 \\
& \Rightarrow \quad 2 x-10 y-4=4 \sqrt{(x-3)^2+(y+2)^2} \\
& \Rightarrow \quad x-5 y-2=2 \sqrt{(x-3)^2+(y+2)^2} \\
& \text { On squaring, }(x-5 y-2)^2=4\left[(x-3)^2+(y+2)^2\right]
\end{aligned}$
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