Problem 10 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A line passing through the point \(P(4,2)\), meets the \(x\)-axis and \(y\)-axis at \(A\) and \(B\), respectively. If \(O\) is the origin, then locus of the centre of the circum circle of \(\triangle O A B\) is (A) \(x^{-1}+y^{-1}=2\) (B) \(2 x^{-1}+y^{-1}=1\) (C) \(x^{-1}+2 y^{-1}=1\) (D) \(2 x^{-1}+2 y^{-1}=1\)

Step-by-Step Solution

Answer

The locus is given by \(x^{-1} + y^{-1} = 2\).

Step 1: Understand the geometry

First, let's determine how the points are located. The line passes through the point \( P(4, 2) \). It intersects the x-axis at point \( A \) where the \( y \)-coordinate is zero, and the y-axis at point \( B \) where the \( x \)-coordinate is zero.

Step 2: Equation of the line

Since the line passes through point \( P(4,2) \), the equation of the line can be written in the point-slope form. The slope \( m \) is \(-(\frac{b}{a})\), where \( a \) and \( b \) are the intercepts on the x and y axes, respectively. So, the equation is \( y-2 = m(x-4) \).

Step 3: Identify the intercepts A and B

On the x-axis (at \( y = 0 \)), the point \( A = (a, 0) \). Substituting \( y = 0 \) in the line equation gives \( a = \frac{8-b}{m}\). On the y-axis (at \( x = 0 \)), the point \( B = (0, b) \). Substituting \( x = 0 \) gives \( b = 2 - 4m \).

Step 4: Find ultimate result for intercepts

Using \( m = -\frac{b}{a} \), substitute \( b = 2 - 4m \) into \( a = \frac{8-b}{m} \), eventually leading to the relation \( m = -1 \). Thus, it simplifies both intercepts as symmetric forms \( a = 2 \), \( b = 2 \).

Step 5: Write the locus condition

The center of the circumcircle of \( \Delta OAB \) is located at \( (a/2, b/2) \). If \( a = 2c \) and \( b = 2c \), where \( c \) is the variable, then \( A = (2c, 0) \) and \( B = (0, 2c) \) makes the center at \( (c, c) \). Use the condition for \( c \) that relates \( a \) and \( b \), giving \( \frac{1}{a} + \frac{1}{b} = \frac{1}{x} + \frac{1}{y} = 1 \).

Step 6: Simplify to a final equation

Recognizing that the center locus is symmetric, substitute \( x = a \) and \( y = b \), obtaining \( \frac{1}{x} + \frac{1}{y} = 1 \), keeping in mind the values are symmetric. This represents expression \( x^{-1} + y^{-1} = 2 \) and solution \("A"\) is evaluated.

Key Concepts

Locus

In coordinate geometry, a locus defines a collection of points that satisfy a particular condition or a set of conditions. When we say that "the locus of a point," we are referring to tracing the path that a point takes when it moves according to certain rules.
Exploring the locus allows us to sketch or describe the path geometrically. For example, in many problems involving circles, the locus could be the path at a fixed distance from a given point, known as the circle's center.

In our exercise, the locus of the center of the circumcircle of \( \triangle OAB \) refers to the path that the center takes as the coordinates of points \( A \) and \( B \) change when the line's position varies while passing through a fixed point \( P(4,2) \). Understanding how to find and interpret the locus is essential for solving many problems in coordinate geometry.

Circumcircle

A circumcircle of a triangle is a circle that passes through all three vertices of the triangle. The center of this circle is called the circumcenter, and the radius is known as the circumradius.
The circumcenter can be found as the intersection point of the perpendicular bisectors of the triangle's sides. It provides significant insight into the properties of the triangle, especially when studying paths and circumferences.

In our problem, we are asked to determine the path that the center of the circumcircle (circumcenter) takes as the line through \( P(4, 2) \) intersects the coordinate axes at points \( A \) and \( B \). It requires determining how the circumcenter moves as points \( A \) and \( B \) are positioned, which influences the intercepts on the axes.

Intercept form

The intercept form of the equation of a line is a convenient way to express how a line cuts across the x-axis and y-axis.This form is beneficial as it directly uses the x-intercept \( a \) and y-intercept \( b \) of a line.
The standard intercept form of a line is given by \( \frac{x}{a} + \frac{y}{b} = 1 \), showing the intercepts clearly.

In this exercise, knowing how to find and utilize the x and y intercepts helps to trace the triangle \( \triangle OAB \), its circumcircle, and find the condition for the circumcenter.The intercepts \( a \) and \( b \) help us find point positions, which further assist in setting up and solving the equation for the locus. This involves utilizing the point-slope form initially and transitioning into the intercept form.

Equation of a Line

The equation of a line is a fundamental concept in geometry that represents all the points lying on that line. Various forms of equations such as the point-slope form and the intercept form are often used.
For a given point-slope form, the equation is \( y-y_1 = m(x-x_1) \) where \( m \) represents the slope, and \( (x_1, y_1) \) is a point on the line.

In the given problem, understanding how the equation through a point \( P(4,2) \) is formed helps deduce where it intercepts the axes at points \( A \) and \( B \). This forms the key to determining the intercept form and, consequently, the locus of the circumcenter.Once the line's intercepts are determined, forming the intercept form equation aids in connecting those intercepts with the triangle \( \Delta OAB \) and its circumcircle.