Question

The differential equation of the family of circles passing the origin and having center at the line y = x is:

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

Answer 1

The Correct Option is C

Consider a family of circles that pass through the origin and have their centers located on the line \( y = x \). The general form of such a circle's equation is given by:

\[ (x - a)^2 + (y - a)^2 = r^2, \] where \( (a, a) \) represents the center of the circle lying on the line \( y = x \), and \( r \) denotes the radius.

Step 1: Expand the standard circle equation

Upon expanding \( (x - a)^2 + (y - a)^2 = r^2 \), we get: \[ x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = r^2. \] Simplifying this expression yields: \[ x^2 + y^2 - 2a(x + y) + 2a^2 = r^2. \]

Step 2: Eliminate the parameters \( a \) and \( r \)

Given that the circle passes through the origin, substituting \( x = 0 \) and \( y = 0 \) into the equation results in: \[ 0^2 + 0^2 - 2a(0 + 0) + 2a^2 = r^2 \implies r^2 = 2a^2. \] Substituting this relationship back into the equation, we obtain: \[ x^2 + y^2 - 2a(x + y) = 0. \] Differentiating both sides with respect to \( x \) produces: \[ 2x + 2y \frac{dy}{dx} - 2a\left(1 + \frac{dy}{dx}\right) = 0. \] Rearranging this equation to solve for \( a \): \[ a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}. \] Substituting the expression for \( a \) back into the circle equation yields: \[ (x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy. \] Consequently, the differential equation governing this family of circles is: \[ (x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy. \]