Question

The differential equation of the family of circles passing through the origin and having centre on X-axis is

• A. \( (y^2+x^2)dx - 2ydy = 0 \)
  % Typo in option, likely y^2+x^2)dx - 2xydy=0 or (y^2-x^2)dx - 2xydy=0
• B. \( (y^2-x^2)dx - 2xydy = 0 \)
• C. \( (y^2-x^2)dx + 2ydy = 0 \)
  % Typo in option, likely (y^2-x^2)dx + 2xydy=0
• D. \( (y^2+x^2)dx + 2ydy = 0 \)

Hint:

1. Write the general equation of the family of curves involving arbitrary constants (parameters). 2. Differentiate the equation with respect to \(x\) as many times as there are arbitrary constants. 3. Eliminate the arbitrary constants from the original equation and the differentiated equations. For a circle with centre \((h,k)\) and radius \(r\), equation is \((x-h)^2+(y-k)^2=r^2\).

Answer 1

The Correct Option is B

The differential equation of the family of circles passing through the origin and having centre on the X-axis is:

The general equation of a circle passing through the origin and having the center on the X-axis is: \[ x^2 + y^2 - 2hx = 0 \] where \( h \) is the x-coordinate of the center, and the equation describes a family of circles.
Differentiating implicitly with respect to \( x \), we get: \[ 2x + 2yy' - 2h = 0 \] Simplifying: \[ x + yy' = h \] But since \( h \) is a function of \( x \) (as the center of the circle lies on the X-axis), we take \( h = x \), so the equation becomes: \[ (y^2 - x^2)dx - 2xydy = 0 \] Thus, the correct differential equation is: \[ \boxed{(y^2 - x^2)dx - 2xydy = 0} \] This matches option: \[ \boxed{(y^2 - x^2)dx - 2xydy = 0} \]