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 \( r \) with centers 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, we get:
\[ x^2 + (y^2 - 2ry + r^2) = r^2 \]
This simplifies to:
\[ x^2 + y^2 - 2ry = 0 \]

### Step 3: Differentiate the equation
To eliminate \( r \), we differentiate the equation with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - 2r\frac{dy}{dx} = 0 \]
This gives us:
\[ 2x + 2y\frac{dy}{dx} - 2r\frac{dy}{dx} = 0 \]

### Step 4: Rearrange the differentiated equation
Rearranging the equation, we have:
\[ 2x + 2y\frac{dy}{dx} = 2r\frac{dy}{dx} \]
Dividing through by 2, we get:
\[ x + y\frac{dy}{dx} = r\frac{dy}{dx} \]

### Step 5: Solve for \( r \)
Rearranging further, we can express \( r \) in terms of \( x \), \( y \), and \( \frac{dy}{dx} \):
\[ r = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \]

### Step 6: Substitute \( r \) back into the original equation
Now, we substitute this expression for \( r \) back into the original equation \( x^2 + y^2 - 2ry = 0 \):
\[ x^2 + y^2 - 2\left(\frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}\right)y = 0 \]

### Step 7: Simplify the equation
After substituting and simplifying, we will have an equation involving \( x \), \( y \), and \( \frac{dy}{dx} \).

### Step 8: Determine the order of the differential equation
Since we differentiated the original equation once to eliminate \( r \), the order of the resulting differential equation is 1.

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