JEE Main 2022 (Online) 30th June Morning ShiftQuestion 1
Let $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$, $a,b,c\in\mathbf{R}$, represents a circle with center $(\alpha,\beta)$. Then $\alpha+2\beta$ is equal to :
Differential Equation
Solution
Concept. Written as $M\,dx+N\,dy=0$ the equation is exact; a circle needs no $xy$ term and equal $x^2$, $y^2$ coefficients. Formulas used. (1) $M\,dx+N\,dy=0$ is exact if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$Step 1 - Exact form. $(-ax+by-a)\,dx+(bx+cy+a)\,dy=0$; both partial derivatives in (1) equal $b$. Step 2 - Solution. $$-\frac a2x^2+bxy-ax+\frac c2y^2+ay=k.$$ Step 3 - Circle. $b=0$ and $\frac c2=-\frac a2$, i.e. $c=-a$. Dividing by $-\frac a2$: $$x^2+y^2+2x-2y=k'.$$ Step 4 - Centre. $(-1,1)$, so $\alpha+2\beta=-1+2=1$. Answer: (C)