Question about two lines and a circle in the cartesian plane

Question

Let $L_1$ be the straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle on $x^2+y^2-x+3y=0$ on $L_1$ and $L_2$ are equal, then find the equation for $L_1$

I found the centre of the circle $(\frac 12 , -\frac 32)$

What does intercept even mean here? The circle well intersect the line at two points, so it the length of the chord? Once we get they cleared up we solve the eqautions of the circle and $x+y=1$ to get $$x=1,2; y=0,-1$$ what should I do with this information?

Edit: I solved using the hints given and obtained $y-x=0$ and $x+7y=0$ as the equation. According to the answer, the latter is incorrect and should be $y+7x=0$. Can anyone please verify this?

Answer 1

Intercept on line by a curve is length of chord intercepted. $$ $$ In circles length of chord are equal if distance of chord from centre of circle are equal. $$ $$ Let line $L_1$ be $y=mx$. Distance of $L_1\,\,,\,\,L_2$ from centre is $$\frac{\vert \frac{3}{2}+\frac{m}{2} \vert}{\sqrt{1+m^2}}=\sqrt{2}$$ $$\frac{\vert m+3 \vert}{\sqrt{1+m^2}}=2\sqrt{2}$$ $$m^2+9+6m=8m^2+8$$ $$7m^2-6m-1=0$$ $$\therefore m=-\frac{1}{7} \,\,,\,\,1$$ Hence line $L_1$ is $7y=-x$ , $y=x$.

Answer 2

Straight line $L_2$ has evident intersection points with circle $C$ which are $(1,0)$ and $(2,-1)$ (see figure). Therefore, the intercept length is $\sqrt{2}$.

$L_1$ (assumed non vertical) has equation

$$y=ax.\tag{1}$$

The abscissas of the intersection points of $C$ and $L_1$ are solutions of the the equation obtained by plugging (1) into the equation of the circle. As a result, we have this quadratic equation :

$$(1+a^2)x^2+(3a-1)x=0$$

$x=0$ is a natural solution (point $O$). The other solution is point

$$I=\left(x=\dfrac{1-3a}{1+a^2},y=ax=a\dfrac{1-3a}{1+a^2}\right).$$

Let us express that the square of intercept length OI has to be equal to the square of the other intercept $\sqrt{2}$ :

$$\left(\dfrac{1-3a}{1+a^2}\right)^2+\left(a\dfrac{1-3a}{1+a^2}\right)^2=2.$$

Otherwise said :

$$(1-3a)^2(1+a^2)=2(1+a^2)^2$$

Simplifying by $(1+a^2)\neq 0$, we get the following quadratic :

$$7a^2-6a-1=0$$

with 2 solutions :

$a=1 \ \implies \ y=1x$

$a=-\dfrac{1}{7} \ \implies \ y=-\tfrac{1}{7}x.$

Fig. 1 : $L_2$ in blue, the two solutions for $L_1$ in magenta.