Write order of the differential equation of the family of following curves
`(x-a)^2+(y-b)^2=9`

Answer

To find the order of the differential equation of the family of curves given by the equation \((x-a)^2 + (y-b)^2 = 9\), we will follow these steps:

### Step 1: Understand the given equation
The equation \((x-a)^2 + (y-b)^2 = 9\) represents a family of circles with center \((a, b)\) and radius \(3\).

### Step 2: Differentiate the equation with respect to \(x\)
We differentiate both sides of the equation with respect to \(x\):

\[ \frac{d}{dx}[(x-a)^2] + \frac{d}{dx}[(y-b)^2] = \frac{d}{dx}[9] \]

Using the chain rule, we get:

\[ 2(x-a) + 2(y-b) \frac{dy}{dx} = 0 \]

### Step 3: Simplify the first derivative
Rearranging the equation gives:

\[ 2(x-a) + 2(y-b) \frac{dy}{dx} = 0 \]

Dividing through by \(2\):

\[ (x-a) + (y-b) \frac{dy}{dx} = 0 \]

This can be rewritten as:

\[ \frac{dy}{dx} = -\frac{x-a}{y-b} \]

### Step 4: Differentiate again to find the second derivative
Now we differentiate \(\frac{dy}{dx} = -\frac{x-a}{y-b}\) with respect to \(x\):

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{(y-b)(1) - (x-a)\frac{dy}{dx}}{(y-b)^2} \]

Substituting \(\frac{dy}{dx} = -\frac{x-a}{y-b}\):

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

This simplifies to:

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

### Step 5: Rearranging the second derivative
We can express this as:

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

### Step 6: Identify the order of the differential equation
The highest derivative present in the equation is \(\frac{d^2y}{dx^2}\), which indicates that the order of the differential equation is \(2\).

### Conclusion
Thus, the order of the differential equation for the family of curves \((x-a)^2 + (y-b)^2 = 9\) is:

\[ \text{Order} = 2 \]