The differential equation of the system of circles touching the y-axis at the origin is

Options

A. ` x^(2) +y^(2) -2xy (dy)/(dx) =0`
B. ` x^(2)+y^(2)+2xy (dy)/(dx)=0`
C. `x^(2) -y^(2) +2xy (dy)/(dx)=0`
D. `x^(2) -y^(2) -2xy(dy)/(dx) =0`

Explanation

To find the differential equation of the system of circles touching the y-axis at the origin, we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle that touches the y-axis at the origin can be expressed as:
\[ (x - a)^2 + y^2 = a^2 \]
where \(a\) is the radius of the circle and also the x-coordinate of the center.

### Step 2: Expand the equation
Expanding the equation gives:
\[ (x - a)^2 + y^2 = a^2 \]
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
Now, simplifying this, we get:
\[ x^2 + y^2 - 2ax = 0 \]

### Step 3: Differentiate with respect to \(x\)
Now, we differentiate the equation \(x^2 + y^2 - 2ax = 0\) with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ax) = 0 \]
Using the chain rule for \(y^2\):
\[ 2x + 2y \frac{dy}{dx} - 2a = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 2y \frac{dy}{dx} = 2a - 2x \]
Dividing through by 2:
\[ y \frac{dy}{dx} = a - x \]

### Step 5: Express \(a\) in terms of \(x\) and \(y\)
From the equation \(x^2 + y^2 - 2ax = 0\), we can express \(a\) as:
\[ a = \frac{x^2 + y^2}{2x} \]

### Step 6: Substitute \(a\) back into the differentiated equation
Substituting \(a\) into \(y \frac{dy}{dx} = a - x\):
\[ y \frac{dy}{dx} = \frac{x^2 + y^2}{2x} - x \]
This simplifies to:
\[ y \frac{dy}{dx} = \frac{x^2 + y^2 - 2x^2}{2x} \]
\[ y \frac{dy}{dx} = \frac{y^2 - x^2}{2x} \]

### Step 7: Rearranging to form the differential equation
Multiplying through by \(2x\) gives:
\[ 2xy \frac{dy}{dx} = y^2 - x^2 \]
Rearranging this gives the required differential equation:
\[ y^2 - 2xy \frac{dy}{dx} - x^2 = 0 \]

### Final Differential Equation
Thus, the differential equation of the system of circles touching the y-axis at the origin is:
\[ y^2 - 2xy \frac{dy}{dx} - x^2 = 0 \]

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