Question

The differential equation of all circles which pass through the origin and whose centre lie on Y-axis is

• A. $(x^2 - y^2) \frac{d y}{d x} - 2xy = 0$
• B. $(x^2 + y^2) \frac{d y}{d x} - 2xy = 0$
• C. $(x^2 + y^2) \frac{d y}{d x} + 2xy = 0$
• D. $(x^2 - y^2) \frac{d y}{d x} + 2xy = 0$

Hint:

A neat geometry property connection: for any circle centered on the y-axis and touching the origin, its algebraic equation is $x^2 + y^2 = 2Ky$. Differentiating it quickly and rearranging terms will always lead to the characteristic pattern $(x^2 - y^2)y' - 2xy = 0$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the differential equation representing the family of circles passing through the origin $(0,0)$ with centers situated on the vertical y-axis.

Step 2: Detailed Explanation:
Since the center of the circle lies on the y-axis, its coordinates can be written as $(0, K)$. Because the circle passes through the origin $(0,0)$, the radius of the circle must be exactly equal to the distance from the center to the origin, which is $K$. Let's write out the general equation of this circle family: $$ (x - 0)^2 + (y - K)^2 = K^2 $$ $$ x^2 + y^2 - 2Ky + K^2 = K^2 $$ $$ x^2 + y^2 - 2Ky = 0 \quad \text{--- (Equation 1)} $$ To eliminate the arbitrary constant $K$, let's differentiate Equation 1 with respect to $x$: $$ 2x + 2y\frac{d y}{d x} - 2K\frac{d y}{d x} = 0 $$ Dividing by 2 and isolating the constant $K$: $$ x + y\frac{d y}{d x} = K\frac{d y}{d x} \implies K = \frac{x + y\frac{d y}{d x}}{\frac{d y}{d x}} $$ Now, let's substitute this expression for $K$ back into Equation 1: $$ x^2 + y^2 - 2\left(\frac{x + y\frac{d y}{d x}}{\frac{d y}{d x}}\right)y = 0 $$ Multiply the entire equation by $\frac{d y}{d x}$ to clear the fractional denominator: $$ (x^2 + y^2)\frac{d y}{d x} - 2y\left(x + y\frac{d y}{d x}\right) = 0 $$ $$ (x^2 + y^2)\frac{d y}{d x} - 2xy - 2y^2\frac{d y}{d x} = 0 $$ Combine the common derivative terms together: $$ (x^2 + y^2 - 2y^2)\frac{d y}{d x} - 2xy = 0 $$ $$ (x^2 - y^2)\frac{d y}{d x} - 2xy = 0 $$

Step 4: Final Answer:
The differential equation of the family of circles is $(x^2 - y^2) \frac{d y}{d x} - 2xy = 0$, matching option (A).