Question

Find the differential equation of all circles passing through the origin and having their centres on the x-axis.

Answer 1

Step 1: General Equation of a Circle:
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

Step 2: Apply the Given Conditions:
* Circle passes through the origin (0, 0):
Substituting \((0, 0)\) into the general equation:
\[ (0 - h)^2 + (0 - k)^2 = r^2 \] Simplifying: \[ h^2 + k^2 = r^2 \]

* Center lies on the x-axis:
This means the y-coordinate of the center is zero, so \( k = 0 \).

Step 3: Substitute \( k = 0 \) into the Equation:
Substituting \( k = 0 \) into \( h^2 + k^2 = r^2 \):
\[ h^2 + 0^2 = r^2 \] Simplifying: \[ h^2 = r^2 \]

Step 4: Simplify the General Equation:
Now, substitute \( k = 0 \) and \( r^2 = h^2 \) into the general equation of the circle: \[ (x - h)^2 + (y - 0)^2 = h^2 \] Simplifying: \[ (x - h)^2 + y^2 = h^2 \] Expanding the equation: \[ x^2 - 2xh + h^2 + y^2 = h^2 \] Simplifying: \[ x^2 - 2xh + y^2 = 0 \]

Step 5: Differentiate with Respect to x:
Differentiate the equation \( x^2 - 2xh + y^2 = 0 \) with respect to \(x\): \[ 2x - 2(h \cdot 1 + x \cdot \frac{dh}{dx}) + 2y \cdot \frac{dy}{dx} = 0 \] Simplifying: \[ 2x - 2h - 2x \cdot \frac{dh}{dx} + 2y \cdot \frac{dy}{dx} = 0 \]

Since \(h\) is a parameter, we consider it constant. However, we know that it is a function of \(x\) and \(y\). From the original equation, we have: \[ 2xh = x^2 + y^2 \] Thus: \[ h = \frac{x^2 + y^2}{2x} \] Substituting \(h\) into the differentiated equation: \[ 2x - 2\left(\frac{x^2 + y^2}{2x}\right) - 2x \cdot \frac{d}{dx}\left(\frac{x^2 + y^2}{2x}\right) + 2y \cdot \frac{dy}{dx} = 0 \]

Simplify further: \[ 2x - \frac{x^2 + y^2}{x} - 2x \cdot \left( \frac{4x^2 - 2(x^2 + y^2)}{4x^2} \right) + 2y \cdot \frac{dy}{dx} = 0 \] Further simplification yields: \[ 2x - \frac{x^2 + y^2}{x} - \frac{2x^2 - 2y^2}{2x} + 2y \cdot \frac{dy}{dx} = 0 \] After simplifying: \[ 2x - \frac{2x^2}{x} + 2y \cdot \frac{dy}{dx} = 0 \] Further simplification: \[ 2x - 2x + 2y \cdot \frac{dy}{dx} = 0 \] This simplifies to: \[ 2y \cdot \frac{dy}{dx} = 0 \] Thus: \[ y \cdot \frac{dy}{dx} = 0 \]

Final Answer:
The final result is: \[ x \cdot \frac{dy}{dx} + y = 0 \]