Problem 108 - A Complete Resource Book in Mathematics for JEE Main

Question

The differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is (A) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) [2007] (B) \(x^{2}=y^{2}+3 x y \frac{d y}{d x}\) (C) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\) (D) \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\)

Step-by-Step Solution

Answer

The differential equation is \(y^2 = x^2 + 2xy \frac{dy}{dx}\), option (C).

Step 1: Identify General Equation of Circle

Start with the general equation of a circle: \[(x - h)^2 + y^2 = r^2\]where \(h\) is the x-coordinate of the circle's center and \(r\) is the radius.

Step 2: Condition for Circles Passing through Origin

Since the circle passes through the origin \((0,0)\), substitute these values into the equation:\[(0 - h)^2 + 0^2 = r^2\]which simplifies to:\[h^2 = r^2\] or \[r = h\]. Thus, the equation becomes:\[(x - h)^2 + y^2 = h^2\].

Step 3: Derive Equation of Family of Circles

From the simplification, we have:\[x^2 - 2hx + y^2 = 0\]. This represents the family of circles with centers on the x-axis passing through the origin.

Step 4: Differentiate with Respect to x

Differentiate \[x^2 - 2hx + y^2 = 0\] with respect to \(x\), using implicit differentiation:\[2x - 2h + 2y \frac{dy}{dx} = 0\].

Step 5: Solve for h

Re-arrange the differentiated equation to solve for \(h\):\[h = x + y \frac{dy}{dx}\].

Step 6: Substitute h Back into Circle Family Equation

Substitute the expression for \(h\) back into the family of circles equation:\[h = x + y \frac{dy}{dx}\] into \[x^2 - 2hx + y^2 = 0\].This yields:\[x^2 - 2(x + y \frac{dy}{dx})x + y^2 = 0\].

Step 7: Simplify to Obtain Differential Equation

Simplify the equation:\[x^2 - 2x^2 - 2xy \frac{dy}{dx} + y^2 = 0\],which results in:\[-x^2 + y^2 - 2xy \frac{dy}{dx} = 0\],or \[y^2 = x^2 + 2xy \frac{dy}{dx}\]. This corresponds to option (C).

Key Concepts

Circles

The concept of circles is fundamental to understanding the geometry of shapes. When we talk about circles in mathematics, we refer to a set of points in a plane that are equidistant from a particular point, known as the center.

The standard equation of a circle with a center at \(h, k\) and a radius \(r\) is:

• \( (x-h)^2 + (y-k)^2 = r^2 \)

For circles that have centers on the x-axis, the y-coordinate of the center, \(k\), is zero. Consequently, the equation simplifies to:

• \( (x-h)^2 + y^2 = r^2 \)

When these circles pass through the origin \( (0, 0) \), the equation becomes more specific. By putting in these origin coordinates into the circle equation, we realize:

• \( h^2 = r^2 \) which implies \( r = h \)

Thus, the equation of all circles passing through the origin and having their centers on the x-axis becomes:

• \( (x-h)^2 + y^2 = h^2 \)

Circles play a major role in differential equations, especially when understanding how curves intersect or touch one another.

Implicit Differentiation

Implicit differentiation is a technique used in calculus when dealing with functions that are not explicitly solved for one of the variables.

Unlike standard differentiation, where a function is given as \(y = f(x)\), implicit differentiation deals with equations where \(y\) and \(x\) are intertwined. In such scenarios, each derivative needs to take into account both variables. For example, if we have an equation:

• \( x^2 - 2hx + y^2 = 0 \)

To find the derivative \(\frac{dy}{dx}\), differentiate each term with respect to \(x\) while treating \(y\) as a function of \(x\). This process might involve using the product rule and chain rule.

• Differentiate \(x^2\) to get \(2x\)
• Differentiate \(-2hx\) (noting \(h\) is a function of \(x\)) to get \(-2h\)
• Differentiate \(y^2\) using the chain rule to get \(2y \frac{dy}{dx} \)

Once the entire equation is differentiated, re-arranging provides a derivative that is \(\frac{dy}{dx}\) in terms of \(x, y\) and other constants or functions. It's particularly useful in finding slopes of tangent lines or solving for values in geometry-based differential equations.

Origin

In mathematics, the origin is a fundamental concept representing the point (0, 0) in a Cartesian coordinate system. It serves as the starting point for measuring coordinates on the x and y axes.

When we speak of circles passing through the origin, we mean that the circle includes this point as part of its circumference. In the given problem, we see this idea reflected in how we substitute origin coordinates into the circle's equation to derive further conclusions such as:

• \( h^2 = r^2 \) when solving for the relationship between the center and radius

The origin is essential for simplifications because of its zero value properties.

• Substituting the origin into an equation often simplifies calculations by eliminating terms, providing direct relationships between others.
• It's the reference point from which all other points in the coordinate plane are defined.

Moreover, understanding circles that pass through the origin helps in visualizing how these shapes interact with the coordinate axes, particularly when analyzing symmetry or solving complex differential equations that involve intersecting curves.