The differential equation of the family of circles passing through the origin and having centres on the x-axis is
Understanding the Problem: Differential Equation of Circle Family
The question asks us to find the differential equation that represents a specific group, or family, of circles. The key characteristics of these circles are:
- They all pass through the origin point (0,0).
- The center of each circle lies somewhere on the x-axis.
We need to translate these geometric conditions into an algebraic equation, introduce a parameter to represent the family, and then eliminate this parameter by differentiation to get the differential equation.
Step-by-Step Derivation of the Differential Equation
Step 1: Write the general equation of a circle
The standard equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]Step 2: Apply the given conditions
We are given two conditions:
- Center is on the x-axis: If the center lies on the x-axis, the y-coordinate of the center is 0. So, \(k = 0\). The equation becomes: \[ (x - h)^2 + (y - 0)^2 = r^2 \] \[ (x - h)^2 + y^2 = r^2 \]
- Circle passes through the origin (0,0): Since the point (0,0) is on the circle, it must satisfy the equation. Substitute \(x=0\) and \(y=0\) into the equation from Step 1: \[ (0 - h)^2 + (0)^2 = r^2 \] \[ (-h)^2 = r^2 \] \[ h^2 = r^2 \] Since \(r\) is the radius, \(r \ge 0\). This tells us that \(r\) must be equal to the absolute value of \(h\), or simply \(r^2 = h^2\).
Step 3: Form the equation of the family of circles
Now substitute \(r^2 = h^2\) back into the equation \((x - h)^2 + y^2 = r^2\) from Step 2:
\[ (x - h)^2 + y^2 = h^2 \]Expand the left side:
\[ x^2 - 2xh + h^2 + y^2 = h^2 \]Subtract \(h^2\) from both sides:
\[ x^2 - 2xh + y^2 = 0 \]This is the equation representing the family of circles passing through the origin and having their centers on the x-axis. Here, \(h\) is the single parameter defining each specific circle in the family.
Step 4: Differentiate the equation to eliminate the parameter \(h\)
The equation of the family is \(x^2 - 2xh + y^2 = 0\). To find the differential equation, we differentiate this equation with respect to \(x\). Remember that \(y\) is a function of \(x\) (\(y\) varies as \(x\) varies on the circle), and \(h\) is a constant parameter for a specific circle.
\[ \frac{d}{dx}(x^2 - 2xh + y^2) = \frac{d}{dx}(0) \]Using the rules of differentiation:
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(2xh) + \frac{d}{dx}(y^2) = 0 \] \[ 2x - 2h \frac{d}{dx}(x) + \frac{d}{dx}(y^2) = 0 \] \[ 2x - 2h(1) + 2y \frac{dy}{dx} = 0 \] \[ 2x - 2h + 2y \frac{dy}{dx} = 0 \]We need to eliminate \(h\) from this equation. From the original family equation \(x^2 - 2xh + y^2 = 0\), we can solve for \(2h\):
\[ 2xh = x^2 + y^2 \] \[ 2h = \frac{x^2 + y^2}{x} \]Now substitute this expression for \(2h\) into the differentiated equation \(2x - 2h + 2y \frac{dy}{dx} = 0\):
\[ 2x - \left(\frac{x^2 + y^2}{x}\right) + 2y \frac{dy}{dx} = 0 \]Multiply the entire equation by \(x\) to clear the denominator:
\[ x(2x) - x\left(\frac{x^2 + y^2}{x}\right) + x\left(2y \frac{dy}{dx}\right) = x(0) \] \[ 2x^2 - (x^2 + y^2) + 2xy \frac{dy}{dx} = 0 \] \[ 2x^2 - x^2 - y^2 + 2xy \frac{dy}{dx} = 0 \] \[ x^2 - y^2 + 2xy \frac{dy}{dx} = 0 \]Rearrange the terms to match the format of the given options:
\[ 2xy \frac{dy}{dx} = y^2 - x^2 \]Step 5: Compare with options
Let's look at the derived differential equation:
\[ 2xy \frac{dy}{dx} = y^2 - x^2 \]Comparing this with the provided options:
- \(2{\rm{xy}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{x}}^2} - {{\rm{y}}^2}\) - This is \(2xy \frac{dy}{dx} = -(y^2 - x^2)\). Does not match.
- \(2{\rm{xy}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{y}}^2} - {{\rm{x}}^2}\) - This matches our derived equation exactly.
- \(2{\rm{xy}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{x}}^2} + {{\rm{y}}^2}\) - Does not match.
- \(2{\rm{xy}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{x}}^2} + {{\rm{y}}^2} = 0\) - This is \(2xy \frac{dy}{dx} = -x^2 - y^2\). Does not match.
The derived differential equation matches Option 2.
Summary of Finding the Differential Equation
We started with the conditions defining the family of circles: passing through the origin and having centers on the x-axis. This led to the family equation \(x^2 - 2xh + y^2 = 0\). By differentiating this equation with respect to \(x\) and eliminating the parameter \(h\), we arrived at the differential equation \(2xy \frac{dy}{dx} = y^2 - x^2\). This equation characterizes all circles that meet the specified criteria.
| Key Step | Description | Equation/Result |
|---|---|---|
| 1 | General Circle Equation | \((x-h)^2 + (y-k)^2 = r^2\) |
| 2 | Apply Conditions (Center on x-axis, passes through origin) | \(k=0\), \(r^2=h^2\) |
| 3 | Equation of Family of Circles | \(x^2 - 2xh + y^2 = 0\) |
| 4 | Differentiate w.r.t. \(x\) and Eliminate \(h\) | \(2xy \frac{dy}{dx} = y^2 - x^2\) |
Revision Table: Key Concepts
| Concept | Brief Explanation | Relevance to Problem |
|---|---|---|
| Differential Equation | An equation involving an unknown function and its derivatives. | The goal is to find the DE representing a family of curves. |
| Family of Curves | A set of curves described by an equation with one or more parameters. | The problem defines a specific family of circles. |
| Parameter | A variable in the equation of a family of curves that distinguishes individual members of the family. | In this problem, \(h\) is the parameter representing the x-coordinate of the center. |
| Forming DE from Family | The process of differentiating the family equation and eliminating the parameter(s). The order of the DE equals the number of independent parameters. | We differentiate once as there is one parameter \(h\). |
| Equation of Circle | \((x-h)^2 + (y-k)^2 = r^2\) where \((h,k)\) is center, \(r\) is radius. | Used as the starting point to define the circle family. |
Additional Information: Families of Curves and Differential Equations
A differential equation can be seen as a way to describe a property that is common to every curve in a family. When you form a differential equation from a family of curves, you are essentially finding a relationship between the coordinates \((x, y)\) and the slope \(\frac{dy}{dx}\) (and possibly higher derivatives) that holds true for any point on any curve in that family, independent of the parameter(s).
- If the family of curves has one essential parameter, the resulting differential equation will be of the first order.
- If the family has two essential parameters, the resulting differential equation will be of the second order, requiring two differentiations and elimination of both parameters.
- The process involves differentiating the family equation as many times as there are parameters and then using algebraic methods to eliminate those parameters from the original equation and the differentiated equations.
In this specific problem, the single parameter \(h\) led to a first-order differential equation.
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