QJEEM24Mabbf5: JEE Main 2024 Mathematics - Class 12 JEEM24Mabbf5 Solution: Suppose the solution of the differential equation represents a…

$\begin{aligned}& \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\& \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\& =2 x d x+\alpha x d x-\beta y d x+2 d x \\& \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\& \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \\& \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \\& \because \text { this represents circle passing through origin } \\& \Rightarrow \beta=0 \text { and } 1+\frac{\alpha}{2}=\alpha \\& \Rightarrow \alpha=2\end{aligned}$

$\begin{aligned}& \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \\& x^2+y^2+x-4 y=0 \\& \text { Radius }=\sqrt{\frac{1}{4}+4-0} \\& \quad=\frac{\sqrt{17}}{2}\end{aligned}$