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(B) The equation of the circle is $x^2+y^2-x+3y=0$. The center is $C = (\frac{1}{2}, -\frac{3}{2})$.Let the line $L_1$ passing through the origin be $

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$x-y=0 	ext{ and } x+7y=0$

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(B) The equation of the circle is $x^2+y^2-x+3y=0$. The center is $C = (\frac{1}{2}, -\frac{3}{2})$.Let the line $L_1$ passing through the origin be $y = mx$,or $mx - y = 0$.The line $L_2$ is $x + y - 1 = 0$.Since the intercepts made by the circle on $L_1$ and $L_2$ are equal,the perpendicular distances from the center $C$ to these lines must be equal.The perpendicular distance from $C(\frac{1}{2}, -\frac{3}{2})$ to $L_1$ is $d_1 = \frac{|m(\frac{1}{2}) - (-\frac{3}{2})|}{\sqrt{m^2 + (-1)^2}} = \frac{|\frac{m+3}{2}|}{\sqrt{m^2+1}}$.The perpendicular distance from $C(\frac{1}{2}, -\frac{3}{2})$ to $L_2$ is $d_2 = \frac{|\frac{1}{2} - \frac{3}{2} - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.Equating $d_1^2 = d_2^2$,we get $\frac{(m+3)^2}{4(m^2+1)} = 2$.$(m+3)^2 = 8(m^2+1) \Rightarrow m^2 + 6m + 9 = 8m^2 + 8$.$7m^2 - 6m - 1 = 0$.$(7m+1)(m-1) = 0$,so $m = 1$ or $m = -\frac{1}{7}$.For $m = 1$,$L_1$ is $y = x$ or $x - y = 0$.For $m = -\frac{1}{7}$,$L_1$ is $y = -\frac{1}{7}x$ or $x + 7y = 0$.

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle $x^2+y^2-x+3y=0$ on $L_1$ and $L_2$ are equal,then which of the following equations represent $L_1$?

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