The differential equation of the system of circles touching the y-axis at the origin is
Differential Equation of Circles Touching Y-axis at Origin
We want to find the differential equation for the system of circles that touch the y-axis at the origin (0, 0). Let's analyze the properties of such circles.
A circle touching the y-axis at the origin must pass through the origin (0, 0) and have its center on the x-axis. Let the center of such a circle be at point \((a, 0)\). Since the circle touches the y-axis at the origin, the distance from the center \((a, 0)\) to the y-axis (which is the x-coordinate, |a|) must be equal to the radius of the circle. Thus, the radius is \(|a|\).
The standard equation of a circle with center \((h, k)\) and radius \(r\) is \((x-h)^2 + (y-k)^2 = r^2\).
Substituting the center \((a, 0)\) and radius \(|a|\), the equation of the system of circles is:
\((x - a)^2 + (y - 0)^2 = a^2\)
Simplifying this equation, we get:
\(x^2 - 2ax + a^2 + y^2 = a^2\)
\(x^2 + y^2 - 2ax = 0\)
This is the equation of the family of circles touching the y-axis at the origin, where \(a\) is the arbitrary constant. To find the differential equation, we need to eliminate this constant \(a\).
Forming the Differential Equation
We have the equation: \(x^2 + y^2 - 2ax = 0 \quad (1)\)
We can differentiate this equation with respect to \(x\) to introduce \(\frac{dy}{dx}\):
Differentiating \(x^2\) with respect to \(x\) gives \(2x\).
Differentiating \(y^2\) with respect to \(x\) gives \(2y\frac{dy}{dx}\) (using the chain rule).
Differentiating \(-2ax\) with respect to \(x\) gives \(-2a\) (since \(a\) is a constant with respect to \(x\)).
So, differentiating equation (1) with respect to \(x\), we get:
\(2x + 2y\frac{dy}{dx} - 2a = 0 \quad (2)\)
Now, we need to eliminate \(a\) from equations (1) and (2). From equation (1), we can express \(a\):
\(2ax = x^2 + y^2\)
\(a = \frac{x^2 + y^2}{2x}\)
Substitute this expression for \(a\) into equation (2):
\(2x + 2y\frac{dy}{dx} - 2\left(\frac{x^2 + y^2}{2x}\right) = 0\)
\(2x + 2y\frac{dy}{dx} - \frac{x^2 + y^2}{x} = 0\)
To clear the denominator, multiply the entire equation by \(x\):
\(x(2x) + x\left(2y\frac{dy}{dx}\right) - x\left(\frac{x^2 + y^2}{x}\right) = x(0)\)
\(2x^2 + 2xy\frac{dy}{dx} - (x^2 + y^2) = 0\)
\(2x^2 + 2xy\frac{dy}{dx} - x^2 - y^2 = 0\)
Combine the \(x^2\) terms:
\((2x^2 - x^2) - y^2 + 2xy\frac{dy}{dx} = 0\)
\(x^2 - y^2 + 2xy\frac{dy}{dx} = 0\)
This is the differential equation for the system of circles touching the y-axis at the origin.
Comparing with Options
Let's compare the derived differential equation with the given options:
| Option | Differential Equation |
|---|---|
| 1 | \({x^2} + {y^2} - 2xy\frac{{dy}}{{dx}} = 0\) |
| 2 | \({x^2} + {y^2} + 2xy\frac{{dy}}{{dx}} = 0\) |
| 3 | \({x^2} - {y^2} + 2xy\frac{{dy}}{dx} = 0\) |
| 4 | \({x^2} - {y^2} - 2xy\frac{{dy}}{{dx}} = 0\) |
The derived differential equation \({x^2} - {y^2} + 2xy\frac{{dy}}{dx} = 0\) matches Option 3.
Revision Table: Key Steps for Differential Equation Formation
| Step | Action | Details |
|---|---|---|
| 1 | Identify the family of curves | System of circles touching y-axis at origin. |
| 2 | Write the general equation | \((x-a)^2 + (y-0)^2 = a^2\), simplifies to \(x^2 + y^2 - 2ax = 0\). |
| 3 | Identify the arbitrary constant(s) | Here, \(a\) is the single arbitrary constant. |
| 4 | Differentiate the equation | Differentiate the general equation with respect to \(x\): \(2x + 2y\frac{dy}{dx} - 2a = 0\). |
| 5 | Eliminate the constant(s) | Use the original equation and the differentiated equation to eliminate \(a\). (e.g., express \(a\) from one and substitute into the other). |
| 6 | Simplify to get the differential equation | \({x^2} - {y^2} + 2xy\frac{{dy}}{dx} = 0\). |
Additional Information on Differential Equations of Families of Curves
The process of finding the differential equation of a family of curves involves eliminating the arbitrary constants from the equation of the family. The order of the resulting differential equation is typically equal to the number of independent arbitrary constants in the equation of the family.
- If the equation of the family has one arbitrary constant, the resulting differential equation will be of the first order.
- If the equation has two arbitrary constants, the resulting differential equation will be of the second order, and so on.
In this specific problem, the family of circles is given by \(x^2 + y^2 - 2ax = 0\), which has only one arbitrary constant, \(a\). Therefore, the resulting differential equation is of the first order, involving \(\frac{dy}{dx}\).
Understanding the geometry of the family of curves is crucial. For circles touching the y-axis at the origin, the center must lie on the x-axis. If the circles touched the x-axis at the origin, the center would lie on the y-axis, and the general equation would be \((x-0)^2 + (y-b)^2 = b^2\), leading to a different differential equation.
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