Question

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

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

Answer 1

Answer: (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)
$\Rightarrow x^2+2gx+c=0$
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 yaxis 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 lies on $x^2-y^2=0$