Problem 39 - A Complete Resource Book in Mathematics for JEE Main

Question

The differential equation of the family of circles with fixed radius 5 units and centre on the line \(y=2\) is (A) \((x-2) y^{\prime 2}=25-(y-2)^{2}\) (B) \((y-2) y^{\prime 2}=25-(y-2)^{2}\) (C) \((y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\) (D) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)

Step-by-Step Solution

Answer

The correct option is (C).

Step 1: Understand the Family of Circles

The problem involves finding the differential equation for a family of circles with a fixed radius of 5 units and centers along the line \(y = 2\). Therefore, the center of any circle in this family is \((h, 2)\) where \(h\) is a variable.

Step 2: Write the Equation of a Circle

A circle with center \((h, 2)\) and radius 5 is described by the equation: \[(x - h)^2 + (y - 2)^2 = 25\]

Step 3: Differentiate the Circle's Equation

To obtain the differential equation, differentiate both sides with respect to \(x\). Use implicit differentiation:\[2(x - h) + 2(y - 2) \cdot \frac{dy}{dx} = 0\]

Step 4: Solve for \(dy/dx\)

Rearrange the differentiated equation to solve for \(dy/dx\):\[(y-2)\frac{dy}{dx} = -(x-h)\]\[\frac{dy}{dx} = \frac{h-x}{y-2}\]

Step 5: Eliminate the Parameter \(h\)

From the circle's original equation, express \(x - h\) in terms of known quantities. Rearrange the circle equation:\[(x - h)^2 = 25 - (y-2)^2\]Using the expression \((x - h) = -(y-2)\frac{dy}{dx}\), equate it to the above to eliminate \(h\):\[(-(y-2)\frac{dy}{dx})^2 = 25 - (y-2)^2\]This simplifies to:\[(y-2)^2 (\frac{dy}{dx})^2 = 25 - (y-2)^2\]

Step 6: Identify the Correct Option

This derived equation is option (C):\[(y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\]

Key Concepts

Family of Circles

A family of circles is a collection of circles that share certain properties. In this exercise, we are dealing with circles that have a fixed radius of 5 units. This means every circle in this family will have the same size but can differ in position. The centers of these circles lie on the line defined by the equation \(y = 2\). Therefore, each circle’s center is represented as \((h, 2)\), where \(h\) varies along the line.

• The position of each center is variable, but the size of the circle is constant, defined by the radius.
• This consistency in properties allows us to express these circles as a family or set defined by similar mathematical equations.

This understanding is crucial for writing the equation of any circle in this family and subsequently deriving a differential equation for them.

Implicit Differentiation

Implicit differentiation is a method used to find the derivative of an equation that isn't solved for one variable in terms of another. Many times, as in the case with circles, resolving for \(y\) directly in terms of \(x\) may result in more complex expressions. Instead, we differentiate both sides with respect to \(x\) while keeping the relationship between \(x\) and \(y\) intact.

• Allows us to differentiate equations that involve multiple inter-dependent variables.
• Used here to derive \(\frac{dy}{dx}\) from the circle's equation, maintaining the integrity of the original relation.

Thus, implicit differentiation is a powerful tool when working with complex equations like those describing a circle.

Eliminate Parameter

To find a differential equation involving only \(x\) and \(y\), we sometimes have to eliminate other parameters, such as \(h\) in our problem. The parameter \(h\) represents the x-coordinate of the center of the circle. By eliminating \(h\), we can condense our problem to only two variables of interest.

• Rewriting or rearranging the original equation, we get \((x-h)^2 = 25 - (y-2)^2\).
• This revealed relationship allows us to replace \((x - h)\) with an expression involving only \(x\) and \(y\).

This manipulation simplifies the equation into a form that isn't dependent on unnecessary parameters, ultimately leading to the correct differential equation.