Question

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

• A. $a=b=0$
• B. $a=-b\ne 0$
• C. $a=b\ne 0,h=k$
• D. $a=b\ne 0$

Answer 1

The correct option is B $a=-b\ne 0$

Given that

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

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

Integrating LHS and RHS,

$b\frac{y^2}{2}+ky+c=a\frac{x^2}{2}+hx+c$

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

The general equation of a cirlce 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 cirlce if

$a=-b\ne 0$.