Question

The differential equation of family of circles whose centres lie on X-axis is

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

Hint:

The number of arbitrary constants in the general equation dictates the order of the resulting differential equation. Because a family of circles on the X-axis can vary in both position ($h$) and size ($r$), the equation must be differentiated exactly twice!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the differential equation representing the family of all circles that have their centers strictly on the X-axis.

Step 2: Key Formula or Approach:
The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.
If the center lies on the X-axis, the y-coordinate of the center ($k$) must be 0. Thus, the center is at $(h, 0)$.
The equation reduces to: $(x - h)^2 + y^2 = r^2$.
Since there are two arbitrary constants ($h$ and $r$), we must differentiate the equation twice to eliminate both constants and obtain a second-order differential equation.

Step 3: Detailed Explanation:
Start with the general equation:
$$(x - h)^2 + y^2 = r^2$$ Differentiate both sides with respect to $x$:
$$2(x - h) \cdot 1 + 2y \frac{dy}{dx} = 0$$ Divide by 2:
$$(x - h) + y \frac{dy}{dx} = 0$$ To eliminate $h$, differentiate again with respect to $x$ using the product rule on the second term:
$$1 + \left( y \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{dy}{dx} \right) = 0$$ $$1 + y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$$ Rearranging to match the options:
$$y\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 + 1 = 0$$

Step 4: Final Answer:
The differential equation is $y\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 + 1 = 0$, which corresponds to option (B).