Understanding the Differential Equation for Circles Touching Coordinate Axes

The question asks for the degree of the differential equation that represents all circles touching both the coordinate axes in the first quadrant. Let's break down how to find this differential equation and its degree.

A circle touching both the x-axis and the y-axis in the first quadrant has its center at a point \((a, a)\) for some positive radius \(a > 0\). The radius of such a circle is also equal to \(a\).

The equation of such a circle is given by:

\((x - a)^2 + (y - a)^2 = a^2\)

This equation represents a family of circles, where \(a\) is the single arbitrary constant or parameter. To form a differential equation from this family, we need to eliminate the parameter \(a\) by differentiation.

Forming the Differential Equation

Since there is only one arbitrary constant \(a\), the resulting differential equation will be of order 1. We differentiate the equation of the family of circles with respect to \(x\):

\(\frac{d}{dx} [(x - a)^2 + (y - a)^2] = \frac{d}{dx} [a^2]\)

Using the chain rule, we get:

\(2(x - a) \frac{d}{dx}(x - a) + 2(y - a) \frac{d}{dx}(y - a) = 0\)

\(2(x - a)(1) + 2(y - a) \left(\frac{dy}{dx} - 0\right) = 0\)

\(2(x - a) + 2(y - a) y' = 0\)

Dividing by 2:

\((x - a) + (y - a) y' = 0\)

Now we need to eliminate \(a\) using this differentiated equation and the original equation \((x - a)^2 + (y - a)^2 = a^2\).

From the differentiated equation, we can express \(x - a\) in terms of \(y - a\) and \(y'\):

\(x - a = -(y - a) y'\)

Substitute this expression for \((x - a)\) into the original equation:

\((-(y - a) y')^2 + (y - a)^2 = a^2\)

\((y - a)^2 (y')^2 + (y - a)^2 = a^2\)

Factor out \((y - a)^2\):

\((y - a)^2 ((y')^2 + 1) = a^2\)

We still have \(a\) and \((y - a)\) in the equation. Let's go back to the differentiated equation \((x - a) + (y - a) y' = 0\) and solve for \(a\):

\(x - a + y y' - a y' = 0\)

\(x + y y' = a + a y'\)

\(x + y y' = a(1 + y')\)

\(a = \frac{x + y y'}{1 + y'}\)

Now, substitute this expression for \(a\) back into the equation \((y - a)^2 ((y')^2 + 1) = a^2\). We also need an expression for \((y - a)\):

\(y - a = y - \frac{x + y y'}{1 + y'} = \frac{y(1 + y') - (x + y y')}{1 + y'} = \frac{y + y y' - x - y y'}{1 + y'} = \frac{y - x}{1 + y'}\)

Substitute both \(a\) and \((y - a)\) into \((y - a)^2 ((y')^2 + 1) = a^2\):

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

\(\frac{(y - x)^2}{(1 + y')^2} ((y')^2 + 1) = \frac{(x + y y')^2}{(1 + y')^2}\)

Assuming \(1 + y' \neq 0\), we can multiply both sides by \((1 + y')^2\):

\((y - x)^2 ((y')^2 + 1) = (x + y y')^2\)

This is the differential equation for the family of circles touching both coordinate axes in the first quadrant.

Determining the Degree of the Differential Equation

The degree of a differential equation is the power of the highest order derivative when the equation is written as a polynomial in the derivatives, free from radicals and fractional powers of the derivatives.

Our differential equation is \((y - x)^2 ((y')^2 + 1) = (x + y y')^2\).

The highest order derivative is \(y' = \frac{dy}{dx}\).

Let's look at the powers of \(y'\) in the equation:

• On the left side, \((y - x)^2 ((y')^2 + 1)\) expands to \((y - x)^2 (y')^2 + (y - x)^2\). The term with \(y'\) has \(y'\) raised to the power of 2.
• On the right side, \((x + y y')^2\) expands to \(x^2 + 2x(y y') + (y y')^2 = x^2 + 2xy y' + y^2 (y')^2\). The terms with \(y'\) have \(y'\) raised to the power of 1 and 2.

Combining terms, the highest power of \(y'\) in the equation is 2. The equation is already in a polynomial form in terms of \(y'\) and there are no fractional powers of \(y'\).

Therefore, the degree of the differential equation \((y - x)^2 ((y')^2 + 1) = (x + y y')^2\) is 2.

Revision Table: Key Concepts

Additional Information: Formation of Differential Equations

The general procedure to form a differential equation for a family of curves with \(n\) arbitrary constants is as follows:

1. Write down the equation of the family of curves.
2. Identify the number of independent arbitrary constants, say \(n\). This will be the order of the differential equation.
3. Differentiate the equation \(n\) times with respect to the independent variable (usually \(x\)).
4. Use the original equation and the \(n\) differentiated equations to eliminate the \(n\) arbitrary constants.
5. The resulting equation, free from arbitrary constants and involving derivatives, is the required differential equation.
6. Once the differential equation is formed, determine its order and degree. The order is the highest derivative present. The degree is the power of that highest derivative after clearing fractions and radicals involving derivatives.

In this specific case, we had one constant \(a\), so we differentiated once and eliminated \(a\) using the original equation and the first derivative.

The resulting differential equation \((y - x)^2 ((y')^2 + 1) = (x + y y')^2\) is a first-order differential equation because the highest derivative is \(y' = \frac{dy}{dx}\). Its degree is 2 because the highest power of \(y'\) is 2.