Question

The differential equation of the family of all circles of radius 'a' is

• A. $y_1y_2 + (1+y_1^2)=a$
• B. $(1+y_1^2)^3 = a^2y_2^2$
• C. $1+y_1^2 = y_2^2+a^2$
• D. $y_2^2+1 = y_1^2+a^2$

Hint:

The differential equation of a family of curves can often be found from a defining geometric property. For a family of circles with a constant radius, the key property is that their radius of curvature is constant and equal to the radius of the circle.

Answer 1

The Correct Option is B

Step 1: Relate the differential equation to a geometric property.
The family of all circles of a fixed radius 'a' has a constant radius of curvature equal to 'a'. The formula for the radius of curvature, $\rho$, for a curve given by $y=f(x)$ is: \[ \rho = \frac{[1 + (y')^2]^{3/2}}{|y''|}. \] where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$.

Step 2: Apply the condition for the family of circles.
For any circle in the family, the radius of curvature is constant and equal to its radius, 'a'. So, we set $\rho = a$. \[ a = \frac{[1 + (y')^2]^{3/2}}{|y''|}. \]

Step 3: Rearrange the equation to match the options.
Move $|y''|$ to the other side: \[ a |y''| = [1 + (y')^2]^{3/2}. \] Square both sides to eliminate the square root and the absolute value: \[ a^2 (y'')^2 = ([1 + (y')^2]^{3/2})^2. \] \[ a^2 (y'')^2 = [1 + (y')^2]^3. \]

Step 4: Rewrite the equation using the notation $y_1$ and $y_2$.
Using the standard notation $y_1 = y'$ and $y_2 = y''$, the differential equation is: \[ a^2 y_2^2 = (1+y_1^2)^3. \] This matches option (B).