The order of the differential equation of all circle of radius r, having centre on y-axis and passing through the origin, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the order of the differential equation of all circles of radius \( r \) having their center on the y-axis and passing through the origin, we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center at \( (0, r) \) and radius \( r \) is given by:
\[ (x - 0)^2 + (y - r)^2 = r^2 \]
This simplifies to:
\[ x^2 + (y - r)^2 = r^2 \]

### Step 2: Expand the equation
Expanding the equation gives:
\[ x^2 + (y^2 - 2yr + r^2) = r^2 \]
This simplifies to:
\[ x^2 + y^2 - 2yr + r^2 - r^2 = 0 \]
Thus, we have:
\[ x^2 + y^2 - 2yr = 0 \]

### Step 3: Eliminate the parameter \( r \)
To find the differential equation, we need to eliminate the parameter \( r \). We can rearrange the equation to express \( r \):
\[ 2yr = x^2 + y^2 \]
This gives:
\[ r = \frac{x^2 + y^2}{2y} \]

### Step 4: Differentiate to eliminate \( r \)
Now, we differentiate both sides with respect to \( x \):
Using implicit differentiation:
\[ \frac{d}{dx}(2yr) = \frac{d}{dx}(x^2 + y^2) \]
Using the product rule on the left side:
\[ 2y \frac{dr}{dx} + 2r \frac{dy}{dx} = 2x + 2y \frac{dy}{dx} \]
Rearranging gives:
\[ 2y \frac{dr}{dx} + 2r \frac{dy}{dx} - 2y \frac{dy}{dx} = 2x \]
Dividing through by 2:
\[ y \frac{dr}{dx} + r \frac{dy}{dx} - y \frac{dy}{dx} = x \]

### Step 5: Identify the order of the differential equation
The resulting equation involves \( r \) and its derivatives. Since we have one parameter \( r \) and its first derivative \( \frac{dr}{dx} \), the order of the differential equation is 1.

### Conclusion
Thus, the order of the differential equation of all circles of radius \( r \) having their center on the y-axis and passing through the origin is:
\[ \text{Order} = 1 \]

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