The differential equation of the family of circles passing through the origin and having centre at t

Q: Why do we only differentiate once? I thought differential equations usually involve higher-order derivatives.

That is a fantastic observation! The order of a differential equation is strictly determined by the number of independent arbitrary constants in the family of curves. Here, our family of circles is defined by a single parameter, $a$. Therefore, we only need one derivative to eliminate it. If we had two parameters, we would indeed need a second-order differential equation. Always count your constants before you start!

Q: Can we use the general form $x^2 + y^2 + 2gx + 2fy + c = 0$?

You absolutely could, but it would be like using a sledgehammer to crack a nut! That general form has three parameters ($g, f, c$). Our specific problem constraints—that the circle passes through the origin ($c=0$) and the center lies on $y=x$ ($g=f=-a$)—reduce it to a single-parameter problem. Always look for ways to simplify the equation using the given constraints before you start calculating. It saves time and prevents errors.

Q: Why did the $a^2$ terms cancel out so perfectly?

That is the sheer beauty of the geometry! Because the circle is constrained to pass through the origin, the radius is intrinsically linked to the center's coordinates. The algebra is simply reflecting the geometric reality that the circle is 'locked' into a specific relationship with the origin. When you see terms canceling out like that, it is a sign that you have correctly captured the geometric essence of the problem.

Question

The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :

Options:

(A) $(x^2 - y^2 + 2xy) dx = (x^2 - y^2 - 2xy) dy$

(B) $(x^2 + y^2 + 2xy) dx = (x^2 + y^2 - 2xy) dy$

(C) $(x^2 + y^2 - 2xy) dx = (x^2 + y^2 + 2xy) dy$

(D) $(x^2 - y^2 + 2xy) dx = (x^2 - y^2 + 2xy) dy$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane. We are looking for a family of circles defined by two specific geometric constraints.

First, their centers lie on the line $y=x$. This means for any circle in our family, if the $x$-coordinate of the center is $a$, the $y$-coordinate must also be $a$. Thus, our center is $C(a, a)$.

Second, every single circle in this family is anchored to the origin, $O(0, 0)$. To write the equation of a circle, we need the center and the radius $r$.

The radius is the distance from the center $(a, a)$ to the origin $(0, 0)$. Using the distance formula:

$$r = \sqrt{(a - 0)^2 + (a - 0)^2} = \sqrt{2a^2} = a\sqrt{2}$$

### The Algebraic Setup

We use the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$. Substituting our center $(a, a)$ and our radius $r = a\sqrt{2}$, we obtain:

$$(x - a)^2 + (y - a)^2 = (a\sqrt{2})^2$$

Expanding this expression, we get:

$$(x^2 - 2ax + a^2) + (y^2 - 2ay + a^2) = 2a^2$$

The $a^2$ terms on the left sum to $2a^2$, which cancels perfectly with the $2a^2$ on the right. We are left with the elegant equation:

$$x^2 + y^2 - 2ax - 2ay = 0$$

To isolate the parameter $a$, we rearrange the terms:

$$x^2 + y^2 = 2a(x + y)$$

### The Calculus of Elimination

Since we have one arbitrary constant $a$, we must differentiate exactly once with respect to $x$ to eliminate it. Differentiating $x^2 + y^2 = 2a(x + y)$ with respect to $x$ yields:

$$2x + 2yy' = 2a(1 + y')$$

Here, $y'$ denotes $\frac{dy}{dx}$. Dividing by 2, we simplify the expression to:

$$x + yy' = a(1 + y')$$

Solving for $a$, we find the key expression:

$$a = \frac{x + yy'}{1 + y'}$$

### Final Synthesis

Now, we substitute this expression for $a$ back into our simplified circle equation $x^2 + y^2 = 2a(x + y)$:

$$x^2 + y^2 = 2 \left( \frac{x + yy'}{1 + y'} \right) (x + y)$$

Multiplying both sides by $(1 + y')$ to clear the fraction, we get:

$$(x^2 + y^2)(1 + y') = 2(x + yy')(x + y)$$

Expanding both sides results in:

$$x^2 + x^2y' + y^2 + y^2y' = 2(x^2 + xy + xyy' + y^2y')$$

$$x^2 + x^2y' + y^2 + y^2y' = 2x^2 + 2xy + 2xyy' + 2y^2y'$$

Grouping all terms containing $y'$ on one side and the remaining terms on the other:

$$x^2y' + y^2y' - 2xyy' - 2y^2y' = 2x^2 + 2xy - x^2 - y^2$$

Simplifying the coefficients, we arrive at the final differential equation:

$$(x^2 - y^2 - 2xy)y' = x^2 - y^2 + 2xy$$

Replacing $y'$ with $\frac{dy}{dx}$, the governing equation is:

$$(x^2 - y^2 + 2xy) dx = (x^2 - y^2 - 2xy) dy$$

**This is the differential equation that governs every single circle in our family.**

Animated Solution for Mathematics - Differential Equations: The differential equation of the family of circles passing through the origin and having centre at the line $y = x$ is :

Select Answer:

Visualized Solution

Visualize the Geometry

Determine the Radius

Standard Equation of Circle

Expand and Simplify

Isolate the Parameter

Differentiate w.r.t. $x$

Atomic Compute: Differentiation

Solve for $a$

Substitution Step

Cross Multiplication

Expand Both Sides

Group $y^{'}$ Terms

Simplify the Coefficients

Final Differential Equation

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Algebraic Setup

The Calculus of Elimination

Final Synthesis

Similar Questions

The differential equation of all circles passing through the origin and having their centres on the $x$ -axis is

The differential equation for the family of circle $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

The differential equation of the family of circles passing through the points $(0, 2)$ and $(0, - 2)$ is

The differential equation of the family of circles with fixed radius 5 units and centre on the line $y = 2$ is

Consider the family of all circles whose centers lie on the straight line $y = x$ . If this family of circle is represented by the differential equation $P y^{''} + Q y^{'} + 1 = 0$ , where $P, Q$ are functions of $x, y$ and $y^{'}$ , then which of the following statements is (are) true?

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is

The differential equation of the family of curves, $x^{2} = 4 b (y + b)$ , $b \in R$ , is:

The differential equation which represents the family of curves $y = c_{1} e^{c_{2} x}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is

The differential equation satisfied by the system of parabolas $y^{2} = 4 a (x + a)$ is:

Let $y = f (x)$ be a curve passing through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves.