Q82 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

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

  • A
    $8 x^2+2 x y+8 y^2+27 x+27 y+45=0$
  • B
    $4 x^2+x y+4 y^2-27 x-27 y+90=0$
  • Correct answer
    C
    $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$
  • D
    $8 x^2-2 x y+8 y^2-27 x-27 y+45=0$

Solution

Let coordinate of variable point $P(h, k)$. Given that, coordinate of $A(1,2), B(2,1)$ and $$ \begin{aligned} & |P A-P B|=3 \\ & \Rightarrow \sqrt{(h-1)^2+(k-2)^2}-\sqrt{(h-2)^2+(k-1)^2}=3 \\ & \Rightarrow \sqrt{(h-1)^2+(k-2)^2}=3+\sqrt{(h-2)^2+(k-1)^2} \end{aligned} $$ Squaring both side $$ \begin{aligned} & \Rightarrow(h-1)^2+(k-2)^2 \\ & =9+(h-2)^2+(k-1)^2+6 \sqrt{(h-2)^2+(k-1)^2} \\ & \Rightarrow h^2+1-2 h+k^2+4-4 k \\ & =9+h^2+4-4 h+k^2+1-2 k \\ & +6 \sqrt{(h-2)^2+(k-1)^2} \\ & \Rightarrow 2 h-2 k-9=6 \sqrt{(h-2)^2+(k-1)^2} \\ & \end{aligned} $$ Again, squaring both sides $$ \begin{aligned} & \Rightarrow(2 h-2 k-9)^2=36\left[(h-2)^2+(k-1)^2\right] \\ & \Rightarrow 4 h^2+4 k^2+81-8 h k+36 k-36 h \\ & =36\left[h^2+4-4 h+k^2+1-2 k\right] \\ & \Rightarrow 4 h^2+4 k^2+81-8 h k+3 k-36 h \\ & =36 h^2+180-144 h+36 k+36-72 k \\ & \Rightarrow 32 h^2+32 k^2+8 h k-108 h-108 k+99=0 \end{aligned} $$ So, locus of $p(h, k)$ is $$ 32 x^2+32 y^2+8 x y-108 x-108 y+99=0 $$

Hence, the correct answer is option: 'C'.

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