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JEE Main 2022 (28 July Shift 2)
LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The differential equation of the family of circles passing through the points and is

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Visualized Solution

Given Points on the Y-Axis

  • Given points: and .
  • These points lie on the y-axis.

Family of Circles

  • We need to find the differential equation for all circles passing through and .

Locus of the Center

  • The center must lie on the perpendicular bisector of the segment joining and .
  • The perpendicular bisector of this segment is the x-axis ().

Coordinates of the Center

  • Let the center be .

Radius of the Circle

  • Radius

General Equation of the Circle

  • Standard form:
  • Substitute center and :

Expanding the Equation

  • Expand the squared term:

Simplifying the Equation

  • Cancel from both sides:
  • Rearrange to isolate the term with :

Differentiating w.r.t.

  • Differentiate with respect to :

Isolating the Parameter

  • From our simplified equation:

Eliminating the Parameter

  • Substitute into the differentiated equation:

Final Algebraic Simplification

  • Multiply the entire equation by :
  • Rearrange terms to match the options:

The Sigma Insight: Formation of Differential Equations

Solution Diagram

The Geometry of Families

A Journey into Differential Equations
Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are uncovering the hidden structure of a family of curves.
When we talk about a 'family of circles,' we are essentially looking at a dynamic system. Imagine a set of circles, all breathing and expanding, yet all anchored to two specific points in space. Our mission is to find the mathematical 'DNA' that governs this entire family—the differential equation.

Phase 1

The Geometric Insight
Let us begin by grounding ourselves in the Cartesian plane. We are given two points: and .
Take a moment to visualize these. They sit on the y-axis, perfectly symmetric about the origin. Now, imagine drawing a circle that passes through both. You could draw a tiny circle that barely connects them, or a gargantuan circle that looks almost like a straight line.
Here is the crucial geometric realization: for any circle passing through these two points, the line segment acts as a chord. A fundamental theorem of geometry tells us that the center of a circle must lie on the perpendicular bisector of any chord.
Since our chord lies on the y-axis, its perpendicular bisector is the x-axis. Therefore, the center of every single circle in our family must lie on the x-axis. We can denote this center as , where is our arbitrary parameter. As changes, the circle shifts, but it always remains anchored to our two points.

Phase 2

Constructing the Equation
Now that we have the center , we need the radius . The radius is simply the distance from the center to any point on the circle. Let us use point .
Using the distance formula, we have:
This simplifies beautifully to:
With the center and the radius squared, we can write the standard equation of the circle:
Substituting our value for , we get:
This is the equation of our family. But we are not done. We need a differential equation, which means we need to eliminate the parameter . Let us expand the left side to see what we are working with:
Notice the elegance of the math here? The terms appear on both sides. We can subtract them, leaving us with a much cleaner expression:
Rearranging this to isolate the term with our parameter , we get:

Phase 3

The Calculus of Elimination
We are now at the threshold of the final step. We have an equation that describes the family, but it still contains the parameter .
To find the differential equation, we must differentiate with respect to . This is the process of 'filtering' out the constant by observing how the curve changes.
Differentiating both sides of with respect to :
Applying the power rule and the chain rule, we obtain:
We are almost there! We have an expression for . Now, we look back at our previous equation, . From this, we can see that:
By substituting this expression for into our differentiated equation, we effectively eliminate the parameter entirely:

The Final Flourish

To reach the standard form, we multiply the entire equation by :
Finally, moving all terms to one side, we arrive at the differential equation of our family of circles:
Look at what we have achieved. We started with a geometric intuition about circles and chords, and through the systematic application of algebra and calculus, we distilled that infinite family of circles into a single, precise differential equation. This is the power of mathematics—the ability to describe complex, infinite systems with a single, elegant line of code.

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