Let L₁ be a straight line passing through the origin and L₂ be the straight line x + y = 1. If the i

Q: Why can't we just calculate the length of the chord directly? It seems more intuitive.

You absolutely could, but in the JEE Advanced arena, time is your most precious resource. Calculating the length of a chord requires finding the radius, then using Pythagoras' theorem, which is a multi-step process. By using the 'equal distance' property, you bypass the radius calculation entirely and jump straight to the algebraic relationship. It is not just faster; it is more elegant.

Q: Why did we assume the equation of $L_1$ is $y = mx$?

That is a great observation! We assume $y = mx$ because the problem explicitly states that $L_1$ passes through the origin $(0,0)$. Any non-vertical line passing through the origin takes the form $y = mx$. If the line were vertical ($x=0$), the distance formula would still work, but the slope $m$ would be undefined. In this specific problem, our quadratic equation for $m$ yields finite values, confirming that our slope-intercept assumption covers the valid solutions.

Q: Why do we keep the modulus sign $|...|$ in the distance formula?

The modulus is crucial because the distance from a point to a line is a physical length, which must be positive. When you substitute the coordinates into the line equation, the result can be negative depending on which side of the line the point lies. By keeping the modulus, you ensure you are dealing with the absolute distance. When you square both sides later, the modulus naturally disappears, but you must respect it during the setup phase to avoid sign errors.

Question

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 can represent $L_1$?

Options:

(A) $x + y = 0$

(B) $x - y = 0$

(C) $x + 7y = 0$

(D) $x - 7y = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Circle

The given equation of the circle is $x^2 + y^2 - x + 3y = 0$. To reveal its geometric properties, we complete the square for both $x$ and $y$ terms:

$$(x^2 - x + \frac{1}{4}) + (y^2 + 3y + \frac{9}{4}) = \frac{1}{4} + \frac{9}{4}$$

This simplifies to the standard form:

$$(x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{10}{4}$$

From this, we identify the center of the circle, $C$, at the coordinates $(\frac{1}{2}, -\frac{3}{2})$.

### The Geometric Principle

The problem states that lines $L_1$ and $L_2$ create equal intercepts on the circle. In geometry, these intercepts are chords of the circle.

A fundamental property of circles is that **equal chords are equidistant from the center**. Therefore, the perpendicular distance from the center $C$ to line $L_1$ must be equal to the perpendicular distance from $C$ to line $L_2$.

### Calculating the Distance to $L_2$

The line $L_2$ is given by $x + y - 1 = 0$. We use the perpendicular distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ for the point $(\frac{1}{2}, -\frac{3}{2})$:

$$d_2 = \frac{|\frac{1}{2} - \frac{3}{2} - 1|}{\sqrt{1^2 + 1^2}}$$

$$d_2 = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$$

### Solving for $L_1$

Line $L_1$ passes through the origin, so its equation is $y = mx$, or $mx - y = 0$. We require the distance from $C(\frac{1}{2}, -\frac{3}{2})$ to $L_1$ to be $\sqrt{2}$:

$$\frac{|m(\frac{1}{2}) - (-\frac{3}{2})|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$$

$$\frac{|\frac{m+3}{2}|}{\sqrt{m^2 + 1}} = \sqrt{2}$$

Squaring both sides to solve for $m$:

$$\frac{(m+3)^2}{4(m^2+1)} = 2$$

$$(m+3)^2 = 8(m^2 + 1)$$

### Final Calculation

Expanding the equation above leads to:

$$m^2 + 6m + 9 = 8m^2 + 8$$

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

Factoring the quadratic equation:

$$(7m + 1)(m - 1) = 0$$

This yields two possible slopes: $m = 1$ and $m = -\frac{1}{7}$. Substituting these back into the line equation $y = mx$, we obtain the final equations for $L_1$:

**$x - y = 0$** and **$x + 7y = 0$**.

Select Answer:

Visualized Solution

Visualizing the Problem

Finding Circle's Center

Distance to $L_{2}$

Calculating $d_{2}$

Equation of $L_{1}$

Setting up the Distance Equation

Squaring Both Sides

Expanding the Equation

Forming the Quadratic Equation

Solving for Slope $m$

Final Equations of $L_{1}$

The Sigma Insight: Intercepts Made by a Circle

Analyzing the Circle

The Geometric Principle

Calculating the Distance to $L_{2}$

Solving for $L_{1}$

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Problem

Finding Circle's Center

Distance to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​

Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​