Question

If the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x}{y}=\frac{\mathrm{a}}{y}\) where a is constant, represents a family of circles then the radius of the circle is \(\qquad\)

• A. \(\mathrm{a}+2\mathrm{c}\), where c is the constant of integration
• B. \(\sqrt{\mathrm{a}^{2}+2\mathrm{c}}\), where c is the constant of integration
• C. \(\mathrm{a}^{2}+2\mathrm{c}\), where c is the constant of integration
• D. \(\sqrt{a+c}\), where \(c\) is the constant of integration

Hint:

Separate the variables, integrate, and complete the square to read off the circle's radius.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Multiply the equation by \(y\) to separate variables. The result should integrate to the equation of a circle.

Step 2: Separate and Integrate
\(y\,dy + x\,dx = a\,dx\)
Integrating: \(\dfrac{y^2}{2}+\dfrac{x^2}{2}=ax+c\), so \(x^2+y^2-2ax=2c\).

Step 3: Complete the Square
\((x-a)^2+y^2=a^2+2c\)
This is a circle with centre \((a,0)\).

Step 4: Read the Radius
Radius \(=\sqrt{a^2+2c}\), where \(c\) is the constant of integration.

Final Answer:
The radius of the circle is \(\sqrt{a^2+2c}\). \[ \boxed{\text{(B) }\sqrt{\mathrm{a}^{2}+2\mathrm{c}}\text{, where c is the constant of integration}} \]