Question

Locus of the centre of the circle which always passes through the fixed point $(a,0)$ and $(-a,0)$, where $a\ne 0$, is

• A. $x=1$
• B. $x+y=6$
• C. $x+y=2a$
• D. $x=0$

Answer 1

The correct option is D $x=0$

Let given points be $A=(a,0),B=(-a,0)$ and the center be $C(h,k)$, then
$C{A}^2=C{B}^2\Rightarrow (h-a{)}^2+k^2=(h+a{)}^2+k^2\Rightarrow 4ah=0\Rightarrow h=0$

$\therefore$ The required locus is $x=0$