Question

The lengths of the intercepts made by any circle on the coordinates axes are equal if the centre lies on the line (s) represented by

• A. x+y+1=0
  
• B. x+y=1
  
• C. $x^2-y^2=0$
  
• D. x-y=1

Answer 1

The correct option is C

$x^2-y^2=0$

Let the equation of any circle be
$x^2+y^2+2gx+2fy+c=0\left(i\right)$
For intercept made by the circle on x-axis, put y=0 in (i)
$\Rightarrow x^2+2gx+c=0\left(ii\right)$
If $x_1,x_2$ are roots of (ii), then length of the intercept on x-axis is
$|x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}=2\sqrt{g^2-c}$
Similarly, length of the intercept of the y-axis is $2\sqrt{f^2-c}$
Since, the lengths of these intercepts are equal
$\sqrt{g^2-c}=\sqrt{f^2-c}$
$\Rightarrow g^2=f^2=(-g{)}^2=(-f{)}^2$
Therefore, centre on $x^2-y^2=0$