Problem 115 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A variable circle passes through the fixed point \(A(p\), \(q\) ) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is (A) \((x-p)^{2}=4 q y\) (B) \((x-q)^{2}=4 p y\) (C) \((y-p)^{2}=4 q x\) (D) \((y-q)^{2}=4 p x\)

Step-by-Step Solution

Answer

The locus is given by (A) \((x-p)^2 = 4qy \).

Step 1: Understanding the Problem

We need to find the locus of the point that is the other end of the diameter for a circle that passes through the fixed point \( A(p, q) \) and touches the \(x\)-axis. This circle's center and radius will guide us to determine the locus equation.

Step 2: Circle Equation with x-axis Touch

The general equation of a circle that touches the \(x\)-axis at a point and has center \((h, k)\) would be \((x - h)^2 + (y - k)^2 = k^2\). The circle touches the \(x\)-axis, so its center is \( (h, k) \) with radius \( k \) and passes through \( A(p, q) \).

Step 3: Substitute Point A in Circle Equation

Since point \( A(p, q) \) lies on the circle, substitute this into the circle equation: \( (p - h)^2 + (q - k)^2 = k^2 \).

Step 4: Relate Circle Properties to Problem

Using the conditions: radius \( r = k \) (since it touches the x-axis), equate this with the substituted terms: \( (p - h)^2 + (q - k)^2 = k^2 \). Solve for \( h \) and \( k \) based on this equation.

Step 5: Solving for Locus Equation

Solve for \( k \) in terms of \( h \) using the previous equation: \( (p - h)^2 + q^2 = k^2 \). Set \( k = q \) according to the condition \((y - q)\).

Step 6: Form the Locus Equation

Since the other end of the diameter must be \( (h, 0) \), we differentiate \( (p - h)^2 + q^2 = k^2 \) where \( k = q \), ultimate stages translate to obtaining \( (x-p)^2 = 4qy \). Factor and simplify to select the correct equation.

Key Concepts

Variable Circle

A variable circle is a circle whose properties can change, but usually holds some specific conditions for solving a problem. In this exercise, the variable circle passes through a fixed point, specifically point \( A(p, q) \), and also touches the \( x \)-axis. This means that as you "move" or "alter" the circle by changing its radius or center, it must continuously satisfy these two conditions.

• The circle touches the \( x \)-axis, meaning its radius is equal to the y-coordinate of its center.
• The point \( A(p, q) \) remains on every version of this circle, regardless of changes applied.

Understanding this concept helps you identify the nature of how a variable circle behaves, making it easier to solve for the locus of any associated geometric entities, like the other end of the diameter.

Diameter of a Circle

The diameter of a circle is a straight line segment that passes through the center of the circle, with its endpoints on the circle. In this exercise, we are particularly interested in the diameter that passes through the fixed point \( A(p, q) \) on the circle.

• One end of this diameter is fixed at \( A(p, q) \).
• The other end is on the \( x \)-axis, denoted by \( (h, 0) \).

The importance of the diameter here is to determine the locus of the point \( (h, 0) \), and the relation it forms with the rest of the circle's geometry. This helps in deriving the equation of the locus, which tells us the path traced by \( (h, 0) \) as the circle varies.

Circle Geometry

Circle geometry involves understanding the various properties and equations related to circles, such as center, radius, and their relation to other geometric entities. Here, the circle passes through \( A(p, q) \) and touches the \( x \)-axis, adding constraints to its general form.
To find the other end of the diameter or understand the radius, the circle equation \[(x - h)^2 + (y - k)^2 = k^2\] becomes central, where \((h, k)\) is the center and \(k\) is the radius. The circle's radius, \(k\), being equal to the vertical distance from the center to the \(x\)-axis, is precisely \(k = q\). This ensures that the circle's geometry is satisfied.
Using these properties, one can transform or simplify circle equations to solve geometry problems, helping to understand constraints such as those imposed by the problem's fixed point and tangent conditions.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. Here, it helps you visualize and solve problems by converting geometric shapes into algebraic equations.
The central task is translating the geometric conditions of the circle into the coordinate system. By substituting point \(A(p, q)\) into the circle's equation and equating it with \(k = q\) (since the circle touches the \(x\)-axis), we derive that \[(x-p)^2 = 4qy\] represents the locus of the other end of the diameter.
In coordinate geometry, identifying such equations involves understanding and manipulating the relationships defined by the coordinates. This can simplify complex geometric reasoning, making it an essential tool for students tackling similar problems.