$A(2,3), B(-1,1)$ are two points. If $P$ is a variable point such that $\angle A P B=90^{\circ}$ then locus of $P$ is

• Correct answer
  A

  $x^2+y^2-x-4 y+1=0$
• B

  $x^2+y^2+x+4 y-1=0$
• C

  $x^2+y^2-x+4 y-1=0$
• D

  $x^2+y^2+x-4 y+1=0$

Solution

$\mathrm{A}(2,3), \mathrm{B}(-1,1)$ and $\angle \mathrm{APB}=90^{\circ}$ Using Pythagoras theorem
$\begin{aligned} & \mathrm{AB}^2=\mathrm{AP}^2+\mathrm{BP}^2 \\ & \Rightarrow(2+1)^2+(3-1)^2=(x-2)^2+(y-3)^2+(x+1)^2 \\ & \Rightarrow+(y-1)^2 \\ & \Rightarrow 13=2 x^2+2 y^2-2 x-8 y+15 \\ & \Rightarrow x^2+y^2-x-4 y+1=0\end{aligned}$

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