Question

Let L₁ be a line passing through the origin and L₂ be the line x + y = 1. If the intercepts made by the circle x²+y²-x+3y = 0 on L₁ and L₂ are equal then the equation of L₁ can be

• A. x+y=0
• B. x-y=0
• C. x + 7y = 0
• D. x-7y=0

Answer 1

Answer: (B), (C)

x-y=0, x + 7y = 0

Answer 2

The equation of the circle is $x^2 + y^2 - x + 3y = 0$. The center of the circle is $C = (1/2, -3/2)$ and the radius squared is $R^2 = 5/2$. The distance from the center to the line L₂ ($x+y-1=0$) is $d_2 = \sqrt{2}$. The length of the intercept on L₂ is $2\sqrt{R^2 - d_2^2} = \sqrt{2}$. Let L₁ be $y=mx$ or $mx-y=0$. The distance from the center to L₁ is $d_1 = \frac{|m+3|}{2\sqrt{m^2+1}}$. Since the intercepts are equal, $d_1 = d_2 = \sqrt{2}$. Squaring this, $d_1^2 = 2$. Thus, $\frac{(m+3)^2}{4(m^2+1)} = 2$, which simplifies to $7m^2 - 6m - 1 = 0$. Factoring gives $(7m+1)(m-1) = 0$, so $m=1$ or $m=-1/7$. If $m=1$, L₁ is $x-y=0$. If $m=-1/7$, L₁ is $x+7y=0$. Both are valid equations for L₁.