Answer:
[tex]y^2+x^2=9[/tex]
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:
[tex]\displaystyle m_1=\frac{y-0}{x-3}=\frac{y}{x-3}[/tex]
The slope of the line passing through P(x,y) and B(-3,0) is:
[tex]\displaystyle m_2=\frac{y-0}{x+3}=\frac{y}{x+3}[/tex]
Substituting in the equation:
[tex]\displaystyle \frac{y}{x-3}\cdot\frac{y}{x+3}=-1[/tex]
Operating:
[tex]\displaystyle \frac{y^2}{(x-3)(x+3)}=-1[/tex]
[tex]y^2=-(x-3)(x+3)[/tex]
[tex]y^2=-(x^2-9)[/tex]
[tex]y^2=-x^2+9[/tex]
[tex]\boxed{y^2+x^2=9}[/tex]
The locus is a circle centered at the origin with radius 3
