The differential equation of all the circles passing through the point `(0,2) and (0,-2)` , is

Options

A. `2xyy'+(x^2+y^2+4)=0`
B. `2xyy'+(x^2-y^2+4)=0`
C. `2xyy'+(-x^2+y^2+4)=0`
D. `2xyy'+(x^2+y^2-4)=0`

Answer

To find the differential equation of all the circles passing through the points (0, 2) and (0, -2), we can follow these steps:

### Step 1: General Equation of a Circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants.

### Step 2: Substitute the Points
Since the circle passes through the points (0, 2) and (0, -2), we can substitute these points into the circle's equation.

**For the point (0, 2):**
\[ 0^2 + 2^2 + 2g(0) + 2f(2) + c = 0 \implies 4 + 4f + c = 0 \implies c = -4 - 4f \quad \text{(1)} \]

**For the point (0, -2):**
\[ 0^2 + (-2)^2 + 2g(0) + 2f(-2) + c = 0 \implies 4 - 4f + c = 0 \implies c = -4 + 4f \quad \text{(2)} \]

### Step 3: Equate the Two Expressions for \(c\)
From equations (1) and (2):
\[ -4 - 4f = -4 + 4f \]
This simplifies to:
\[ -4f - 4f = 0 \implies -8f = 0 \implies f = 0 \]

### Step 4: Substitute \(f\) Back to Find \(c\)
Substituting \(f = 0\) back into either equation for \(c\):
\[ c = -4 - 4(0) = -4 \]

### Step 5: Substitute \(f\) and \(c\) Back into the Circle Equation
Now we can substitute \(f\) and \(c\) back into the general equation of the circle:
\[ x^2 + y^2 + 2gx + 0 + (-4) = 0 \implies x^2 + y^2 + 2gx - 4 = 0 \]

### Step 6: Differentiate the Circle Equation
To find the differential equation, we differentiate the circle equation with respect to \(x\):
\[ \frac{d}{dx}(x^2 + y^2 + 2gx - 4) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} + 2g = 0 \]
Rearranging this, we find:
\[ 2y \frac{dy}{dx} = -2x - 2g \implies y \frac{dy}{dx} = -x - g \]

### Step 7: Eliminate the Parameter \(g\)
From the circle equation \(x^2 + y^2 + 2gx - 4 = 0\), we can express \(g\):
\[ g = \frac{4 - x^2 - y^2}{2} \]
Substituting this back into the differentiated equation:
\[ y \frac{dy}{dx} = -x - \frac{4 - x^2 - y^2}{2} \]
Simplifying gives:
\[ y \frac{dy}{dx} = -x - 2 + \frac{x^2 + y^2}{2} \]

### Step 8: Rearranging to Get the Final Differential Equation
Multiplying through by 2 to eliminate the fraction:
\[ 2y \frac{dy}{dx} = -2x - 4 + x^2 + y^2 \]
Rearranging gives:
\[ x^2 + y^2 - 2y \frac{dy}{dx} - 2x - 4 = 0 \]

Thus, the differential equation of all circles passing through the points (0, 2) and (0, -2) is:
\[ x^2 + y^2 - 2y \frac{dy}{dx} - 2x - 4 = 0 \]