Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

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 answer is (A).

Step 1: Write the Equation of the Circle

Consider a circle with radius 5 units and a center at some point \((h, 2)\). The general equation for this circle is \((x - h)^2 + (y - 2)^2 = 25\).

Step 2: Differentiate with Respect to x

Differentiate both sides of \((x - h)^2 + (y - 2)^2 = 25\) with respect to \(x\). This gives \(2(x - h) + 2(y - 2)\frac{dy}{dx} = 0\).

Step 3: Simplify the Derivative Equation

Solve for \(\frac{dy}{dx}\) by rearranging the differentiated equation: \(2(y - 2)\frac{dy}{dx} = -2(x - h)\), which simplifies to \(\frac{dy}{dx} = \frac{-(x - h)}{(y - 2)}\).

Step 4: Express h in Terms of y and x

To obtain a family of equations, note that \(h\) is arbitrary since the circle's center is on the line \(y = 2\). Substitute \(h = x + (y - 2)\frac{dy}{dx}\) into \((y - 2)\frac{dy}{dx} = -(x - h)\).

Step 5: Form the Required Differential Equation

Substitute \(\frac{dy}{dx}\) back into the circle equation to eliminate \(h\): \((y - 2)^2 \left( \frac{dy}{dx} \right)^2 = (x - h)^2\). With \((x-h)^2 = 25 - (y-2)^2\), we derive the differential equation: \((x - h)^2 y^{\prime 2} = 25 - (y - 2)^2\).

Step 6: Match with Given Options

Since we expressed \( x-h \) in terms of \( y \), and \( h \) is eliminated, the correct option that matches is (A): \((x-2) y^{\prime 2}=25-(y-2)^{2}\).

Key Concepts

Family of Circles

A family of circles is a set of circles that are described by an equation with certain parameters varying. In this exercise, we are looking at circles that share a common property—their centers all lie on a specific line.
The problem specifically deals with circles of radius 5, and centers on the line \(y = 2\). This means every circle has the same distance from its center to any point on its circumference: 5 units as the radius.

• Circular centers align along \(y = 2\), ensuring all circles share their horizontal position.
• Each circle's general equation can be written as \((x-h)^2 + (y-2)^2 = 25\).
• The parameters vary; in this case, \(h\), which allows infinite circles along the line.

Understanding this helps us identify the broad characteristics all circles in the family must fulfill, making subsequent calculations and simplifications contextually grounded.

Derivative Calculation

To find a differential equation that describes the family of circles, we must differentiate the original circle equation. Differentiation gives us a representation of the slope of the tangent to the circle at any given point.
Starting with \((x-h)^2 + (y-2)^2 = 25\), we need to apply implicit differentiation:

• Differentiate \((x-h)^2\) to get \(2(x-h)\), which reflects changes in \(x\).
• Differentiate \((y-2)^2\) using the chain rule to get \(2(y-2)\frac{dy}{dx}\).

This process of implicit differentiation is crucial as it gives us \[ 2(x-h) + 2(y-2)\frac{dy}{dx} = 0 \].
By rearranging for \(\frac{dy}{dx}\) to find the derivative, you prepare for the elimination of the arbitrary parameter \(h\) in the final steps, simplifying the problem to fit a specific form.

Equation Simplification

After differentiating the circle equation, our goal is to eliminate the arbitrary parameter \(h\). This step is necessary to derive a differential equation that truly reflects the characteristics of the entire family of circles, without depending on any specific center.
Post differentiation, you have the expression \(\frac{dy}{dx} = \frac{-(x-h)}{(y-2)}\). The simplification involves expressing \(h\) in terms of other components within the circle’s family equation:

• Use the property that \(h\) cannot be determined independently because it resides on the constant line \(y=2\).
• Substitute \(h = x + (y - 2)\frac{dy}{dx}\) into the derivative equation.

This substitution helps eliminate \(h\), yielding the relation: \[(y-2)^2 \left( \frac{dy}{dx} \right)^2 = 25 - (y-2)^2\].
Therefore, matching this to an option, you find that the differential equation \((x-2) y^{\prime 2}=25-(y-2)^{2}\) appropriately describes the family, aligning with Option (A). Simplifying equations is pivotal in making them practical and meaningful in describing entire families or sets of geometric shapes.