1. Understand the condition:
The tangent at any point \( (x, y) \) has \( x \)-intercept \( 2x \) and \( y \)-intercept \( 2y \).
2. Find the equation of the tangent:
The tangent line passes through \( (2x, 0) \) and \( (0, 2y) \). Its slope is:
\[ m = \frac{0 - 2y}{2x - 0} = -\frac{y}{x} \]
Equation of the tangent:
\[ Y - y = -\frac{y}{x}(X - x) \]
3. Find the differential equation:
At \( X = 0 \), \( Y = 2y \):
\[ 2y - y = -\frac{y}{x}(-x) \implies y = y \]
This holds true, so we use the slope condition:
\[ \frac{dy}{dx} = -\frac{y}{x} \]
4. Solve the differential equation:
Separate variables:
\[ \frac{dy}{y} = -\frac{dx}{x} \implies \ln y = -\ln x + C \implies \ln(xy) = C \implies xy = e^C = C' \]
Correct Answer: (A) \( xy = C \)
For a curve $xy = C$, consider a point $(a, b)$ on the curve. The equation of the tangent at this point is derived using implicit differentiation: \[ b \, dx + a \, dy = 0 \implies \frac{dy}{dx} = -\frac{b}{a}. \] The equation of the tangent line is: \[ y - b = -\frac{b}{a}(x - a) \implies y = -\frac{b}{a}x + 2b. \] The $x$-intercept is when $y = 0$: \[ 0 = -\frac{b}{a}x + 2b \implies x = 2a. \] The $y$-intercept is when $x = 0$: \[ y = 2b. \] Thus, the intercepts are $(2a, 0)$ and $(0, 2b)$, which are double the coordinates of the point $(a, b)$. Therefore, the family of curves satisfying the given condition is $xy = C$.