Obtain the differential equation of the family of circles passing through the point (a,0) and (-a,0).

Options

A. `y_(1)(y^(2)-x^(2))+2xy+a^(2)`
B. `y_(1)y^(2)+xy+a^(2)x^(2)=0`
C. `y_(1)(y^(2)-x^(2)+a^(2))+2xy=0`
D. none of these

Answer

To obtain the differential equation of the family of circles passing through the points (a, 0) and (-a, 0), we can follow these steps:

### Step 1: Equation of the Circle
The general equation of a circle with center at (0, h) and radius r can be written as:
\[ x^2 + (y - h)^2 = r^2 \]
Since the circle passes through the points (a, 0) and (-a, 0), we can express the radius in terms of these points.

### Step 2: Distance from Center to Points
The distance from the center (0, h) to the point (a, 0) is:
\[ r = \sqrt{(a - 0)^2 + (0 - h)^2} = \sqrt{a^2 + h^2} \]
Thus, the equation of the circle becomes:
\[ x^2 + (y - h)^2 = a^2 + h^2 \]

### Step 3: Expanding the Circle Equation
Expanding the left-hand side:
\[ x^2 + (y^2 - 2hy + h^2) = a^2 + h^2 \]
This simplifies to:
\[ x^2 + y^2 - 2hy = a^2 \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ x^2 + y^2 - a^2 - 2hy = 0 \]

### Step 5: Differentiating the Equation
Now, we differentiate the equation with respect to x:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(a^2) - \frac{d}{dx}(2hy) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} - 2h \frac{dy}{dx} = 0 \]

### Step 6: Solving for h
Rearranging the equation:
\[ 2x + (2y - 2h) \frac{dy}{dx} = 0 \]
From this, we can isolate h:
\[ h = y + \frac{x}{\frac{dy}{dx}} \]

### Step 7: Substituting h Back
Substituting the value of h back into the rearranged circle equation:
\[ x^2 + y^2 - a^2 - 2(y + \frac{x}{\frac{dy}{dx}})y = 0 \]
This simplifies to:
\[ x^2 + y^2 - a^2 - 2y^2 - \frac{2xy}{\frac{dy}{dx}} = 0 \]

### Step 8: Final Rearrangement
Rearranging gives us:
\[ x^2 - y^2 + a^2 + \frac{2xy}{\frac{dy}{dx}} = 0 \]

### Final Differential Equation
Thus, the differential equation of the family of circles passing through the points (a, 0) and (-a, 0) is:
\[ y^2 - x^2 + a^2 + 2xy \frac{dy}{dx} = 0 \]