Question

The solution of $\frac{dy}{dx}=\frac{ax+h}{by+k}$represents a circle, when

• A. $a=b$
• B. $a=–b\ne 0$
• C. $a=–2b=0$
• D. $a=2b$

Answer 1

The correct option is B

$a=–b\ne 0$

Find the required condition

Given, $\frac{dy}{dx}=\frac{ax+h}{by+k}$

$\Rightarrow (by+k)dy=(ax+h)dx$

On integrating both sides, we get

$$\begin{gathered} \int (by+k)dy=\int (ax+h)dx \\ \Rightarrow b\frac{y^2}{2}+ky+c_1=a\frac{x^2}{2}+hx+c_2 \\ \Rightarrow \frac{b{y}^2+2ky+2c_1}{2}=\frac{a{x}^2+2hx+2c_2}{2} \\ \Rightarrow a{x}^2-b{y}^2+2hx-2ky+2c_2-2c_1=0 \end{gathered}$$

$\Rightarrow a{x}^2-b{y}^2+2hx-2ky+c=0$, where $2c_2-2c_1=c$

The general equation of a circle is given as,

$x^2+y^2+2gx+2fy+c=0$

Upon comparing the obtained equation with the general equation of a circle,

$a{x}^2-b{y}^2+2hx-2ky+c=0$

represents a circle if the coefficient of $x^2\text{and}{y}^2$ are same,

$a=-b\ne 0$

Hence, option (B) is the correct answer.