Question

Find the differential equation of the family of all circles, whose center lies on the x-axis and touches the y-axis at the origin.

• A. \( 2xy \frac{dy}{dx} = y^2 - x^2 \)
• B. \( 2xy \frac{dy}{dx} = x^2 - y^2 \)
• C. \( x^2 + y^2 = 2xy \frac{dy}{dx} \)
• D. \( x^2 + y^2 = 2y \frac{dy}{dx} \)

Hint:

When differentiating implicit equations, apply the chain rule and remember to differentiate each term with respect to \( x \). For a family of circles, center and radius conditions are key.

Answer 1

The Correct Option is A

The equation of a circle centered at \( (h, 0) \) with radius \( h \) is \( (x - h)^2 + y^2 = h^2 \). Expanding this equation yields \( x^2 + y^2 + h^2 - 2hx = h^2 \), which simplifies to \( x^2 + y^2 - 2hx = 0 \). Differentiating implicitly with respect to \( x \), we get \( 2x + 2y \frac{dy}{dx} - 2h = 0 \). This can be rearranged to express \( h \) as \( h = x + y \frac{dy}{dx} \). Substituting this expression for \( h \) back into the simplified equation \( x^2 + y^2 - 2hx = 0 \) gives \( x^2 + y^2 - 2x \left( x + y \frac{dy}{dx} \right) = 0 \). Further expansion results in \( x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \), which simplifies to \( y^2 - x^2 - 2xy \frac{dy}{dx} = 0 \). Rearranging to find the equation of the family of circles, we obtain \( 2xy \frac{dy}{dx} = y^2 - x^2 \).