Problem 29 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(\mathrm{B}\). A circle is drawn through \(\mathrm{A}, \mathrm{B}\) and the origin. Then the sum of perpendicular distances from \(\mathrm{A}\) and \(\mathrm{B}\) on the tangent to the circle at the origin is : (a) \(\frac{\sqrt{5}}{2}\) (b) \(2 \sqrt{5}\) (c) \(\frac{\sqrt{5}}{4}\) (d) \(4 \sqrt{5}\)

Step-by-Step Solution

Answer

The sum of indefinite distances is \(\frac{2\sqrt{5}}{5}\) (not matching any given option).

Step 1: Find the Intercepts

The line \(x + 2y = 1\) meets the x-axis when \(y = 0\). Setting \(y = 0\) in the equation yields \(x = 1\). Therefore, the x-intercept \(A\) is \((1, 0)\).ewlineSimilarly, the line meets the y-axis when \(x = 0\). Setting \(x = 0\) in the equation yields \(2y = 1\) or \(y = \frac{1}{2}\). Therefore, the y-intercept \(B\) is \((0, \frac{1}{2})\).

Step 2: Find the Circle's Equation

The circle passes through the points \((1, 0)\), \((0, \frac{1}{2})\), and the origin \((0, 0)\). Its general equation can be written as: \(x(x - 1) + y(y - \frac{1}{2}) = c\). Since the circle passes through the origin, the equation simplifies to \(x(x - 1) + y(y - \frac{1}{2}) = 0\), which is \(x^2 - x + y^2 - \frac{1}{2}y = 0\).

Step 3: Find the Tangent to the Circle at the Origin

The tangent to the circle at the origin \((0, 0)\) is obtained by differentiating the equation of the circle: \(2x - 1, 2y - \frac{1}{2}\). At the origin, this results in the line \(y = \frac{1}{2}x\). The tangent line's equation can be rewritten as \(x - 2y = 0\).

Step 4: Calculate Perpendicular Distances

The perpendicular distance from a point \((x_1, y_1)\) to a line \(ax + by + c = 0\) is given by the formula \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\). For point \((1, 0)\) (A), the distance to line \(x - 2y = 0\) is \(\frac{|1 \cdot 1 - 2 \cdot 0|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{5}}\). For point \((0, \frac{1}{2})\) (B), the distance is \(\frac{|1 \cdot 0 - 2 \cdot \frac{1}{2}|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{5}}\).

Step 5: Sum the Distances

The sum of the perpendicular distances from \(A\) and \(B\) to the tangent line is \(\frac{1}{\sqrt{5}} + \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}\). This further simplifies to \(\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\).

Key Concepts

Equation of a Line

Understanding the equation of a line is crucial in coordinate geometry. Lines in two-dimensional space can be represented by linear equations such as \(x + 2y = 1\). This equation is in the standard line form \(ax + by = c\), where \(a\), \(b\), and \(c\) are real numbers.
To find where the line meets the axes, we look for intercepts. The **x-intercept** is found by setting \(y = 0\), leading to the equation \(x = 1\). Hence, the point \((1, 0)\) is where the line crosses the x-axis. Similarly, for the **y-intercept**, we set \(x = 0\), which gives \(y = \frac{1}{2}\), so the point \((0, \frac{1}{2})\) is the y-intercept.
Intercepts are foundational in many geometry problems because they provide anchor points that help define the geometry of other shapes, like circles that may intersect or be tangent to the line.

Circle Geometry

Circle geometry involves understanding shapes like circles and their relationships with lines, points, and other geometric figures. In this exercise, a circle passes through three specific points: the x-intercept \((1, 0)\), the y-intercept \((0, \frac{1}{2})\), and the origin \((0, 0)\).
The equation of the circle needs to satisfy all these points, which allows us to find its equation by using the general formula for a circle that is not centered on the origin. Substituting the points into the circle equation helps determine unknown constants, eventually yielding the specific equation \(x^2 - x + y^2 - \frac{1}{2}y = 0\).
Understanding how a circle interacts with given points and lines is a key part of circle geometry. This includes finding tangents at specific points, which are lines that just "touch" the circle at those points without crossing it.

Perpendicular Distance Formula

The perpendicular distance formula allows us to calculate the shortest distance from a point to a line. This is vital in scenarios where we need to determine how far a specific point is from a given line, like finding distances from the intercepts of a line to a tangent.
For any point \((x_1, y_1)\) to a line given by \(ax + by + c = 0\), the perpendicular distance \(d\) is calculated as:

• \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\)

In this exercise, applying the formula calculates the distances from points \((1, 0)\) and \((0, \frac{1}{2})\) to the line \(x - 2y = 0\). Both calculations yield distances of \(\frac{1}{\sqrt{5}}\). These distances combined give the total sum of distances, which is crucial when calculating features like total path lengths or when proving geometric properties.
Understanding this formula deepens the comprehension of spatial relationships and is often used in verifying whether two lines are parallel or perpendicular based on their intersections with other lines or shapes.