Question

The differential equation of the family of circles passing through \((0,0)\) and having centre on \(x\)-axis is

• A. \(2xy\dfrac{dy}{dx}+x^2-y^2=0\)
• B. \(\left(\dfrac{dy}{dx}\right)^2+y\dfrac{d^2y}{dx^2}+1=0\)
• C. \(xy\dfrac{dy}{dx}+y^2-x^2=0\)
• D. \(\dfrac{dy}{dx}=\dfrac{x+y}{x-y}\)

Hint:

To form a differential equation from a family of curves, first write the general equation with arbitrary constants, then differentiate and eliminate those constants.

Answer 1

The Correct Option is A

Step 1: Set up the family of circles.
Circles through \((0,0)\) with centre on x-axis have centre \((h,0)\) and radius \(h\) (since they pass through origin). Equation: \((x-h)^2+y^2=h^2\Rightarrow x^2-2hx+y^2=0\Rightarrow h=\frac{x^2+y^2}{2x}\).

Step 2: Differentiate to eliminate h.
Differentiate \(x^2-2hx+y^2=0\) w.r.t. \(x\): \(2x-2h+2yy'=0\Rightarrow h=x+yy'\). Substitute back: \(x^2-2(x+yy')x+y^2=0\Rightarrow x^2-2x^2-2xyy'+y^2=0\Rightarrow y^2-x^2=2xyy'\). Rearranging: \[\boxed{2xy\dfrac{dy}{dx}+x^2-y^2=0}\]