Question

If $A(3,0),B(-3,0)$ then the locus of the point $P$ such that ${PA}^2+{PB}^2=18$ is

• A. $x^2+y^2=0$
• B. $x^2+y^2=8$
• C. $x^2+y^2=9$
• D. $x^2+y^2=3$

Answer 1

The correct option is A $x^2+y^2=0$

Let $(x,y)$ be a point

$\therefore (x-3{)}^2+y^2+(x+3{)}^2+y^2=18$

$\therefore \left(\right(PA{)}^2+(PB{)}^2=18)$

$\therefore x^2-6x+9+y^2+x^2+6x+9+y^2=18$

$\therefore 2(x^2+y^2)+18=18$

$\therefore x^2+y^2=0$