Problem 7 - A Complete Resource Book in Mathematics for JEE Main

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 with \(A B\) as a diameter is (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) none of these

Step-by-Step Solution

Answer

The equation of the circle is (B) \(x^{2}+y^{2}-x-y=0\).

Step 1: Identify the intercepts of the circle and the line

To find the intercepts where the circle \(x^2 + y^2 - 2x = 0\) and the line \(y = x\) intersect, substitute \(y = x\) into the equation of the circle. This gives \(x^2 + x^2 - 2x = 0\), which simplifies to \(2x^2 - 2x = 0\).

Step 2: Solve for x to find intercept points

Factor the equation \(2x^2 - 2x = 0\) as \(2x(x - 1) = 0\). This yields the solutions \(x = 0\) and \(x = 1\). Thus, the points of intersection are \((0, 0)\) and \((1, 1)\).

Step 3: Identify points A and B

Based on the intersection points, \(A\) is \((0, 0)\) and \(B\) is \((1, 1)\). These points are the ends of the diameter of the new circle we need to find.

Step 4: Find the equation of the new circle using diameter AB

The equation of a circle with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) as a diameter is given by \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\). Substituting \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (1, 1)\), you get \((x - 0)(x - 1) + (y - 0)(y - 1) = 0\), simplifying to \(x^2 - x + y^2 - y = 0\).

Step 5: Determine the correct equation from the options

The simplified equation \(x^2 + y^2 - x - y = 0\) matches the option (B), thus the correct answer is (B).

Key Concepts

Circle and Line Intersect

When studying the intersection of a circle and a line, we examine how these two geometric figures can meet at certain points. This concept is fundamental in geometry as it combines properties of both circles and straight lines.

To find where a line intersects a circle, you replace the line's equation into the circle's equation, which is a key step. For the example given, the line has the equation \( y = x \) and the circle \( x^2 + y^2 - 2x = 0 \). Replace \( y \) in the circle's equation:

• Substitute: \( y = x \).
• This yields: \( x^2 + x^2 - 2x = 0 \).
• Which simplifies to: \( 2x^2 - 2x = 0 \).

Then, you solve this resulting equation to find the \( x \)-values of the intersection points, which pairs with \( y = x \) give all the points where the line touches the circle. In this case, the intersection points are \((0, 0)\) and \((1, 1)\). These points are significant for forming elements like a diameter if needed.

Diameter of a Circle

The diameter of a circle is a line segment that passes through the center, with both ends on the circle's boundary. It's essential because it provides information about the circle's size and helps define other aspects such as its radius (half the diameter).

Given points \(A(0, 0)\) and \(B(1, 1)\), these are the endpoints of a diameter. Understanding the geometry, if you have endpoints of a diameter on the circle, you can directly formulate the equation of that circle from these two points.

• Use the formula: \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\).
• Substitute: \((x_1, y_1) = (0, 0), (x_2, y_2) = (1, 1)\).
• Simplified, it's: \( x^2 - x + y^2 - y = 0 \).

This equation characterizes the new circle having the diameter \(AB\). Knowing this helps when matching the given options to find the correct one.

Solving Quadratic Equations

Quadratic equations often appear in problems involving circles and lines, and solving them is key to finding necessary solutions. The general form is \( ax^2 + bx + c = 0 \), and there are several methods to solve it, such as factoring, using the quadratic formula, or completing the square.

In the problem, after substituting the line's equation into the circle's equation, you arrive at \( 2x^2 - 2x = 0 \), a quadratic equation. Here, you can apply factoring:

• Factor out: \( 2x(x - 1) = 0 \).
• This results in solutions: \( x = 0 \) and \( x = 1 \).

Each solution corresponds to an intersection point of the circle and line, which are fundamental in determining aspects like diameter endpoints or verifying intersections. Mastering these methods allows students to tackle more complex problems confidently.