The differential equation of the family of circles passing the origin and having center at the line y = x is:
- \((x^2 - y^2 + 2xy)dx = (x^2 - y^2 + 2xy)dy\)
- \((x^2 + y^2 + 2xy)dx = (x^2 + y^2 - 2xy)dy\)
- \((x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy\)
- \((x^2 + y^2 - 2xy)dx = (x^2 + y^2 + 2xy)dy\)
The Correct Option is C
Solution and Explanation
The family of circles passes through the origin and has centers on the line \( y = x \). The general equation of such a circle is:
\[ (x - a)^2 + (y - a)^2 = r^2, \] where \( (a, a) \) is the center of the circle on the line \( y = x \), and \( r \) is the radius.
Step 1: Expand the circle equation
Expanding \( (x - a)^2 + (y - a)^2 = r^2 \): \[ x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = r^2. \] Simplifying: \[ x^2 + y^2 - 2a(x + y) + 2a^2 = r^2. \]
Step 2: Eliminate parameters \( a \) and \( r \)
Since the circle passes through the origin, substitute \( x = 0 \) and \( y = 0 \) into the equation: \[ 0^2 + 0^2 - 2a(0 + 0) + 2a^2 = r^2 \implies r^2 = 2a^2. \] Thus, the equation becomes: \[ x^2 + y^2 - 2a(x + y) = 0. \] Differentiating both sides with respect to \( x \): \[ 2x + 2y \frac{dy}{dx} - 2a\left(1 + \frac{dy}{dx}\right) = 0. \] Rearranging to isolate \( a \): \[ a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}. \] Substitute \( a \) back into the circle equation: \[ (x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy. \] Thus, the differential equation of the family of circles is: \[ (x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy. \]
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