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

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 \(\Delta O A B\) 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 correct option is (A) \( x^{-1} + y^{-1} = 2 \).

Step 1: Understanding Line Equation

The line passes through point \( P(4, 2) \) and cuts the x-axis at point \( A(x, 0) \) and the y-axis at point \( B(0, y) \). Since \( A \) lies on the x-axis and \( B \) on the y-axis, we need to determine their coordinates in terms of a general line equation.

Step 2: Find Equation of the Line

The equation of a line passing through \( P(4, 2) \) can be written in the intercept form: \[ \frac{x}{a} + \frac{y}{b} = 1 \] where \( a \) and \( b \) are the x-intercept and y-intercept respectively. Since the line goes through \( P(4, 2) \), substitute \( (4, 2) \) into the equation: \( \frac{4}{a} + \frac{2}{b} = 1 \). (Equation 1)

Step 3: Identify Points A and B

From the intercept form, \( A(a, 0) \) and \( B(0, b) \). Thus \( A \) and \( B \) depend solely on the intercepts \( a \) and \( b \). This means \( b = \frac{2}{1 - \frac{4}{a}} \) from rearranging Equation 1.

Step 4: Find Coordinates of Center of Circumcircle

The center of the circumcircle of \( \Delta OAB \), given that \( \angle OAB \) is a right angle with \( BA \perp OA \), lies at the midpoint of the hypotenuse \( AB \). This midpoint \( M \) is \( \left( \frac{a}{2}, \frac{b}{2} \right) \).

Step 5: Finding Locus Equation

We need to express \( M \)'s coordinates in terms of \( x \) and \( y \) to find a relation. Since \( a = 2x \) and \( b = 2y \) from midpoint, we substitute \( x = \frac{a}{2} \) and \( y = \frac{b}{2} \) into the Equation 1: \( \frac{2x}{a} + \frac{2y}{b} = 1 \). This simplifies to: \( \frac{1}{x} + \frac{1}{y} = 2 \).

Step 6: Compare with Given Options

After simplifying, we find the locus equation \( \frac{1}{x} + \frac{1}{y} = 2 \) corresponds to option (A). Compare this respective form to all given options.

Key Concepts

Line Equation

The line equation is fundamental in geometry and algebra. In its simplest form, a line can be represented by the formula \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. However, there are multiple forms of a line equation, and sometimes a more suitable form is the intercept form when dealing with problems involving points on the axes.

For lines cutting both the x-axis and y-axis, the intercept form is useful, characterized as \( \frac{x}{a} + \frac{y}{b} = 1 \). Here, \( a \) is the x-intercept, and \( b \) is the y-intercept.

This form makes it easy to analyze intersections with axes by solving simpler equations. In our problem, substituting point \( P(4, 2) \) into this form gives \( \frac{4}{a} + \frac{2}{b} = 1 \). This allows us to find coordinates of intersection points \( A \) and \( B \).

Line equations are powerful tools for finding relationships between different geometric elements like intersections, distances, and directions.

Circumcircle

A circumcircle or circumscribed circle is a unique circle that passes through all three vertices of a triangle. The center of this circle is known as the circumcenter, and it can be positioned either inside or outside the triangle, depending on the triangle's type.

For right-angled triangles, the circumcenter is at the midpoint of the hypotenuse. This is because the circumcircle's radius equals half the hypotenuse's length. Knowing this property simplifies calculations significantly in such cases.

In our problem, the circumcircle of triangle \(\Delta OAB\) goes through the origin \(O\) and points \(A\) and \(B\), forming a right angle at \(A\) and \(B\). The circumcenter, hence, lies at \((\frac{a}{2}, \frac{b}{2})\), which is the midpoint of the line segment \(AB\).

Understanding circumcircles is crucial in geometry as they help in solving problems involving triangle-centric calculations and locus derivations.

Midpoints

Midpoints are vital in geometry as they help in partition and symmetry problems. The midpoint of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated as \((\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})\).

In this exercise, calculating the midpoint of the line segment \(AB\) allows us to find the circumcenter of the triangle \(\Delta OAB\). This is because, as mentioned, \(AB\) is the hypotenuse of the right triangle formed with the origin, making the midpoint \(\left( \frac{a}{2}, \frac{b}{2} \right)\) the center of the circumcircle.

Employing the midpoint formula here helps derive the locus equation. The points \(A\) and \(B\) also reveal the relationship between the intercepts \(a\) and \(b\), which is used in further computations.

Intercept Form

The intercept form of a line equation, expressed as \( \frac{x}{a} + \frac{y}{b} = 1 \), is especially beneficial in scenarios involving intersections with the coordinate axes.

Each intercept \(a\) and \(b\) can be visualized as the distances from the origin to the intersection on the respective axis. This provides a straightforward understanding of the line's behavior as it cuts through \(A(a,0)\) and \(B(0,b)\).

• The x-intercept \(a\) occurs where the line meets the x-axis, meaning \(y=0\).
• The y-intercept \(b\) occurs where the line meets the y-axis, where \(x=0\).

By recognizing these intersections, the intercept form simplifies analysis and solutions, such as calculating areas under the line or finding other geometric properties like midpoints or circumcenters.

For our exercise, using this form efficiently reveals half-values \(\left(\frac{a}{2}, \frac{b}{2}\right)\) derived from solving the equation, leading to the correct locus representation. It elegantly ties together points and their respective intercepts in problem-solving.