Question

A(3,0) and B(-3,0) are the two fixed points Find the locus of a point P at which AB subtend a right angle at P

Answer 1

Answer:

$y^2+x^2=9$

Step-by-step explanation:

Let's take a general point P(x,y). If there is a right angle at P then the slopes of the lines passing through AP and BP must be perpendicular.

Two lines of slopes m1 and m2 are perpendicular if:

m1*m2=-1

The slope of the line passing through P(x,y) and A(3,0) is:

$\displaystyle m_1=\frac{y-0}{x-3}=\frac{y}{x-3}$

The slope of the line passing through P(x,y) and B(-3,0) is:

$\displaystyle m_2=\frac{y-0}{x+3}=\frac{y}{x+3}$

Substituting in the equation:

$\displaystyle \frac{y}{x-3}\cdot\frac{y}{x+3}=-1$

Operating:

$\displaystyle \frac{y^2}{(x-3)(x+3)}=-1$

$y^2=-(x-3)(x+3)$

$y^2=-(x^2-9)$

$y^2=-x^2+9$

$\boxed{y^2+x^2=9}$

The locus is a circle centered at the origin with radius 3