Question

The ends of hypotenuse of a right angled triangle are $(a,0),(-a,0)$ then the locus of third vertex is:

• A. $x^2-y^2=2a^2$
• B. $x^2-y^2=a^2$
• C. $x^2+y^2=a^2$
• D. $x^2-y^2=4a^2$

Answer 1

Answer: (C)

$x^2+y^2=a^2$

Let the third vertex be $P(x,y)$

Let $C=(a,0)$ and $B=(-a,0)$

Since PCB form a right angled triangle right angled at $P$, then application of Pythagoras theorem, gives us

$(PC{)}^2+(PB{)}^2=(BC{)}^2$

$\left[\right(x-a{)}^2+y^2]+[(x+a{)}^2+y^2]=4a^2$

$\left[\right(x-a{)}^2+(x+a{)}^2)+2y^2=4a^2$

$2x^2+2y^2+2a^2=4a^2$

$2x^2+2y^2=2a^2$

$x^2+y^2=a^2$

This represents a circle.