TS EAMCET · Maths · Straight Lines
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
- A \((x+y+1)^2=4\left[(x-3)^2+(y+2)^2\right]\)
- B \((x-5 y-2)^2=4\left[(x-2)^2+(y-3)^2\right]\)
- C \((x-5 y-2)^2=4\left[(x-3)^2+(y+2)^2\right]\)
- D \((x+y+1)^2=4\left[(x-2)^2+(y-3)^2\right]\)
Answer & Solution
Correct Answer
(C) \((x-5 y-2)^2=4\left[(x-3)^2+(y+2)^2\right]\)
Step-by-step Solution
Detailed explanation
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…
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