Question

Circles are drawn through the point $(3,0)$ to cut an intercept of length $6$units on $x$-axis. The equation of the locus of their centres is

• A. $y(y-6)=0$
• B. $y(x-6)=0$
• C. $x(x-6)=0$
• D. $x(y-6)=0$

Answer 1

Answer: (C)

$x(x-6)=0$

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$
Since, it passes through $(3,0)$
Therefore, $9+0+6g+0+c=0$
$\Rightarrow c=-6g-9$ ....(1)
length of x-intercept $=2\sqrt{g^2-c}=6$
$\Rightarrow g^2+6g+9=9$ [ from (1) ]
$\Rightarrow {(-g)}^2-6(-g)=0$
Therefore, locus of center $x^2-6x=0$
Ans: C