To solve the problem, we need to follow these steps:
### Step 1: Solve the Differential Equation
We start with the given differential equation:
\[
(1 + y^2) dx - xy dy = 0
\]
Rearranging gives:
\[
(1 + y^2) dx = xy dy
\]
Dividing both sides by \(xy(1 + y^2)\):
\[
\frac{dx}{x} = \frac{y}{1 + y^2} dy
\]
### Step 2: Integrate Both Sides
Now we integrate both sides:
\[
\int \frac{dx}{x} = \int \frac{y}{1 + y^2} dy
\]
The left side integrates to:
\[
\ln |x| + C_1
\]
For the right side, we use the substitution \(t = 1 + y^2\), hence \(dt = 2y dy\) or \(dy = \frac{dt}{2y}\):
\[
\int \frac{y}{1 + y^2} dy = \frac{1}{2} \ln(1 + y^2) + C_2
\]
Thus, we have:
\[
\ln |x| = \frac{1}{2} \ln(1 + y^2) + C
\]
### Step 3: Exponentiate to Eliminate Logarithm
Exponentiating both sides gives:
\[
|x| = e^{C} \sqrt{1 + y^2}
\]
Let \(k = e^{C}\), so:
\[
x = k \sqrt{1 + y^2}
\]
### Step 4: Find the Conic Equation
Squaring both sides:
\[
x^2 = k^2 (1 + y^2)
\]
Rearranging gives:
\[
x^2 - k^2 y^2 = k^2
\]
This represents a conic section. Since it passes through the point (1, 0), we substitute \(x = 1\) and \(y = 0\):
\[
1^2 - k^2(0)^2 = k^2 \implies 1 = k^2 \implies k = 1
\]
Thus, the equation of the conic is:
\[
x^2 - y^2 = 1
\]
### Step 5: Identify the Foci of the Conic
The foci of the hyperbola \(x^2 - y^2 = 1\) are located at \((\pm c, 0)\) where \(c = \sqrt{a^2 + b^2} = \sqrt{1 + 1} = \sqrt{2}\). Therefore, the foci are at \((\sqrt{2}, 0)\) and \((- \sqrt{2}, 0)\).
### Step 6: Find the Equation of the Circle
The circle touches the conic at the point \((\sqrt{2}, 1)\) and passes through one of its foci, say \((\sqrt{2}, 0)\). The general equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Here, \((h, k) = (\sqrt{2}, 1)\).
### Step 7: Write the Tangent Equation at the Point of Contact
The tangent line at the point \((\sqrt{2}, 1)\) can be found using the formula for the tangent to the hyperbola:
\[
\sqrt{2}x - y = 1
\]
### Step 8: Substitute into Circle Equation
The equation of the circle can be expressed as:
\[
(x - \sqrt{2})^2 + (y - 1)^2 = r^2
\]
We also know that the circle passes through \((\sqrt{2}, 0)\):
\[
(\sqrt{2} - \sqrt{2})^2 + (0 - 1)^2 = r^2 \implies 0 + 1 = r^2 \implies r^2 = 1
\]
### Step 9: Final Circle Equation
Substituting \(r^2 = 1\) into the circle equation gives:
\[
(x - \sqrt{2})^2 + (y - 1)^2 = 1
\]
Expanding this:
\[
(x^2 - 2\sqrt{2}x + 2) + (y^2 - 2y + 1) = 1
\]
Combining terms:
\[
x^2 + y^2 - 2\sqrt{2}x - 2y + 2 = 0
\]
Rearranging gives:
\[
x^2 + y^2 - 2\sqrt{2}x - 2y + 2 = 0
\]
### Final Answer
The equation of the circle is:
\[
x^2 + y^2 - 2\sqrt{2}x - 2y + 2 = 0
\]
