Question

$A(2,0)$ and $B(-2,0)$ are two given points. Find the equation of locus of point $P$, such that $\angle APB$ is a right angle.

Answer 1

Let $p$ be a point on the locus $AP\perp BP$

Gradient of AP $\left(m_{a}p\right)$

$=(y-0)(x-2)=\frac{y}{x-2}$

Gradient of $BP$ $\left(m_{BP}\right)=\frac{y-0}{x+2}=\frac{y}{x+2}$

Now $AP\perp BP\Rightarrow m_{AP}.m_{BP}=-1$

$\Rightarrow \frac{y}{x-2}.\frac{y}{x+2}-1$

$\Rightarrow y^2=-1(x-2)(x+2)$

$\Rightarrow y^2=-(x^2-4)$

$\Rightarrow y^2+x^2+4=0$

$\Rightarrow x^2+y^2+4=0$ or

$x^2+y^2=4$

Centre of circle $(0,0)r=2$

$x^2+y^2=r^2$

$x^2+y^2=(2{)}^2$

thus $r=2$