Problem 3 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the differential equation representing the family of all circles touching \(\mathrm{x}\)-axis at the origin is \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{g}(\mathrm{x}) \mathrm{y}\), then \(\mathrm{g}(\mathrm{x})\) equals: (a) \(\frac{1}{2} x\) (b) \(2 x^{2}\) (c) \(2 \mathrm{x}\) (d) \(\frac{1}{2} x^{2}\)

Step-by-Step Solution

Answer

The solution for \( g(x) \) is (c) \( 2x \).

Step 1: Recognize the Problem Type

We are tasked with finding the expression for \( g(x) \) in the given differential equation for a family of circles touching the \( x \)-axis at the origin. This involves recognizing that the general form of the equation of a circle is \( (x - a)^2 + (y - b)^2 = r^2 \).

Step 2: Determine the Circle's Parameters

Since circles touch the \( x \)-axis at the origin, the circle is centered at \( (0, r) \) and has radius \( r \). Therefore, the equation of the circle can be written as \( x^2 + (y - r)^2 = r^2 \).

Step 3: Simplify the Circle Equation

Simplify the circle equation: \( x^2 + y^2 - 2yr = 0 \). This is the rearrangement of the circle equation in a form that includes both x and y.

Step 4: Differentiate Implicitly

Differentiate the simplified circle equation with respect to \( x \) to find \( \frac{dy}{dx} \). Differentiating \( x^2 + y^2 - 2yr = 0 \) gives \( 2x + 2y\frac{dy}{dx} - 2\frac{dy}{dx}r = 0 \).

Step 5: Make \( \frac{dy}{dx} \) the Subject

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \). Simplifying, \( 2x = 2\frac{dy}{dx}(r - y) \), thus \( \frac{dy}{dx} = \frac{x}{r-y} \).

Step 6: Substitute and Rearrange the Differential Equation

Return to the given differential equation, \( (x^2 - y^2)\frac{dy}{dx} = g(x)y \). Substitute \( \frac{dy}{dx} = \frac{x}{r-y} \) in, leading to \( (x^2 - y^2)\frac{x}{r-y} = g(x)y \).

Step 7: Simplify to Find \( g(x) \)

Rearrange to solve for \( g(x) \): \( g(x) = \frac{(x^3 - xy^2)}{y(r-y)} \). Since \( y = r \) at the \( x \)-axis point, the denominators simplify. Thus \( g(x) = 2x \), given that the balance of terms must reflect the form of touching circles.

Key Concepts

Family of Circles

A family of circles refers to a group of circles that share a common property. In this exercise, these circles are defined as touching the x-axis at the origin, which means each circle in this family will be tangent to the x-axis at the point (0, 0).
Understanding this helps us identify the characteristics shared by all circles in this family.

• All circles have their tangent point on the x-axis at the origin.
• The centers of these circles lie vertically above or below the x-axis, depending on the radius.
• The radius of each circle equals the y-coordinate of its center, as they must reach down to touch the x-axis.

Comprehending these properties is crucial when setting up the mathematical representation of these circles.

Implicit Differentiation

Implicit differentiation is a technique used to take the derivative of equations where the dependent and independent variables are not easily separated.
It's especially useful in cases like our circle equations, where x and y are related by a more complicated formula rather than an explicit function.
Here's a step-by-step breakdown of implicit differentiation:

• Start with an equation that involves two variables, often mixed together, like our circle's equation.
• Differentiate every term with respect to x, treating y as a function of x (so you'll need to apply the chain rule to terms involving y).
• Once differentiated, you can solve the result for dy/dx, giving you the derivative with respect to x.

In our problem, implicit differentiation of the simplified form \(x^2 + y^2 - 2yr = 0\) helps us find the rate of change of y in relation to changes in x for circles in this family.

Equation of a Circle

The equation of a circle is the mathematical representation of all the points that form the circle.
For a circle centered at a point (a, b) with radius r, the equation is \((x - a)^2 + (y - b)^2 = r^2\).
In this specific exercise, the properties shift slightly due to the unique condition of touching the x-axis at the origin. Here’s how to adapt the equation:

• The center of the circle is at (0, r), leading to the circle's equation being \(x^2 + (y - r)^2 = r^2\).
• Simplifying this equation, we transform it to \(x^2 + y^2 - 2yr = 0\), which is more practical for differentiation.
• This form makes it clear how y and r play roles in defining the circles in this family.

Thus, every circle in our family can be described by its unique equation, all of which stem from these foundational circle properties.