Question

If $A(2,4)$ and $B(6,10)$ are two fixed points and if a point $P$ moves so that $\angle APB$ is always a right angle, then the locus of $P$ is

• A. $x^2+y^2+8x+14y+52=0$
• B. $x^2+y^2-8x-14y-52=0$
• C. $x^2+y^2+8x-14y+52=0$
• D. $x^2+y^2-8x-14y+52=0$

Answer 1

Answer: (D)

$x^2+y^2-8x-14y+52=0$

Given, two points $A(2,4)$ and $B(6,10)$
$\angle APB={90}^o$
$\therefore$ Locus of $P$ is given by $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ where $(x_1,y_1$ and $(x_2,y_2)$ are the given points
$\Rightarrow (x-2)(x-6)+(y-4)(y-10)=0$
$\Rightarrow x^2+y^2-8x-14y+52=0$