Question

Which of the following lines have the intercepts of equal lengths made by the circle $x^2+y^2-2x+4y=0$ is/are

• A. $3x-y=0$
• B. $x+3y=0$
• C. $x+3y+10=0$
• D. $3x-y-10=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $3x-y=0$
B $x+3y=0$
C $x+3y+10=0$
D $3x-y-10=0$

Given circle: $S:x^2+y^2-2x+4y=0$
Centre: $C=(-g,-f)=(1,-2)$
Radius$=\sqrt{g^2+f^2-c}=\sqrt{5}$
Length of intercept made by circle
$=2\sqrt{r^2-d^2}$
Since $r$ is fixed, it depends on $d$ (distance from centre to the line)
$3x-y=0d_1=\frac{\left|3\right(1)+2|}{\sqrt{10}}=\frac{5}{\sqrt{10}}x+3y=0d_2=\frac{|1+3(-2\left)\right|}{\sqrt{10}}=\frac{5}{\sqrt{10}}x+3y+10=0d_3=\frac{|1+3(-2)+10|}{\sqrt{10}}=\frac{5}{\sqrt{10}}3x-y-10=0d_4=\frac{\left|3\right(1)+2-10|}{\sqrt{10}}=\frac{5}{\sqrt{10}}$

$\therefore d_1=d_2=d_3=d_4$
All are having equal intercepts.