Problem 117 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The intercept on the line \(y=x\) by the circle \(x^{2}+y^{2}-2 x\) \(=0\) is \(A B\). Equation of the circle on \(A B\) as a diameter is [2004] (A) \(x^{2}+y^{2}-x-y=0\) (B) \(x^{2}+y^{2}-x+y=0\) (C) \(x^{2}+y^{2}+x+y=0\) (D) \(x^{2}+y^{2}+x-y=0\)

Step-by-Step Solution

Answer

The equation is \(x^2 + y^2 - x - y = 0\) (A).

Step 1: Find Points of Intersection

The given circle's equation is \(x^2 + y^2 - 2x = 0\). The line's equation is \(y = x\). Substitute \(y = x\) into the circle's equation to find the points of intersection.Substituting, we get: \[ x^2 + x^2 - 2x = 0 \]\[ 2x^2 - 2x = 0 \]\[ 2x(x - 1) = 0 \].The solutions to this equation are \(x = 0\) and \(x = 1\). So, the points of intersection are \((0, 0)\) and \((1, 1)\).

Step 2: Determine Diameter Points

The points of intersection \((0, 0)\) and \((1, 1)\) directly give us the endpoints of the diameter of the new circle.

Step 3: Equation of Circle with Given Diameter

The general equation of a circle with diameter endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is:\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \].Plug in the points \((0, 0)\) and \((1, 1)\):\[ (x - 0)(x - 1) + (y - 0)(y - 1) = 0 \]This simplifies to:\[ x(x-1) + y(y-1) = 0 \],\[ x^2 - x + y^2 - y = 0 \].The circle's equation is therefore \(x^2 + y^2 - x - y = 0\).

Step 4: Match with Given Options

Compare the resulting equation \(x^2 + y^2 - x - y = 0\) with the options provided:- (A) \(x^2 + y^2 - x - y = 0\) (Correct)- (B) \(x^2 + y^2 - x + y = 0\)- (C) \(x^2 + y^2 + x + y = 0\)- (D) \(x^2 + y^2 + x - y = 0\)Therefore, option (A) is the correct answer.

Key Concepts

Intercepts

When we talk about intercepts, we usually refer to the points where a curve or a line crosses the axes on a graph. However, in this particular exercise, we're more interested in the intercepts as they relate to another line, specifically the line defined by the equation \(y = x\).

To find these intercepts, or points of interaction, between our circle and the line \(y = x\), we substitute \(y = x\) into the circle's equation \(x^2 + y^2 - 2x = 0\). By doing so, we transform the problem into one with a single variable, making it easier to solve. This substitution gives us \(2x^2 - 2x = 0\), and solving this equation tells us the \(x\)-values where the circle and the line meet: \(x = 0\) and \(x = 1\).

These \(x\)-values lead directly to the points \((0, 0)\) and \((1, 1)\) on the graph, as at these \(x\)-values, \(y\) is equal to \(x\) on the line \(y = x\). This understanding of intercepts between curves is fundamental as it helps us define precise points where two shapes intersect.

Intersection Points

Intersection points are the specific points where two different geometric figures meet or cross each other. When dealing with a circle and a line, as in our exercise, these points can provide meaningful insights into the relationship between the two.

In the problem, after substituting \(y = x\) into the circle's equation, we solve for \(x\), finding \(x = 0\) and \(x = 1\). Each of these values gives us a pair of coordinates where the circle and line intersect: the points \((0, 0)\) and \((1, 1)\).

These intersection points serve two important roles here:

• They are the endpoints of the diameter of a new circle.
• They allow us to derive a new circle's equation by acting as endpoints of a diameter.

Understanding the role of intersection points is crucial because they often determine subsequent calculations, such as when identifying new geometric properties or finding new equations related to specific geometric figures.

Circle Properties

The properties of a circle are essential when forming or analyzing circles in equations. A circle is defined as the set of all points equidistant from a center point. However, in our case, we're interested in forming a new circle using a line segment as its diameter.

Once we've established that the diameter of our new circle is the line segment joining points \((0, 0)\) and \((1, 1)\), we use a special circle property. The equation of any circle with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) of its diameter is given by:

• \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \).

Plugging in \(x_1 = 0, y_1 = 0, x_2 = 1, y_2 = 1\), the equation becomes \( x^2 - x + y^2 - y = 0 \), simplifying to \(x^2 + y^2 - x - y = 0\).

This equation represents the circle with the given diameter and showcases a critical aspect of circle properties: how the relationship between points (in this case, the endpoints of a diameter) can dictate the full description of the circle.