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

`y^2+x^2=9`

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=(y-0)/(x-3)=(y)/(x-3)`

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

`\displaystyle m_2=(y-0)/(x+3)=(y)/(x+3)`

Substituting in the equation:

`\displaystyle (y)/(x-3)\cdot(y)/(x+3)=-1`

Operating:

`\displaystyle (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