Suppose the solution of the differential equation d y/d x=(2+α) x-β y+2/β x-2 α y-(β γ-4 α) represents a circle passing through origin. Then the radius of this circle is :

Mathematics · Differential Equations

JEE Main 2024 — 6 April, Shift 2 — Question 4

Suppose the solution of the differential equation $\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \quad$

represents a circle passing through origin. Then the radius of this circle is :

$\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-y(2 \alpha+\beta)+4 \alpha}$

$\beta x d y-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x-\beta y d x+2 d x$

$\beta(x d y+y d x)-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x+2 d x$

$\beta x y-\frac{(2 \alpha+\beta) y^{2}}{2}+4 \alpha y=\frac{(2+\alpha) x^{2}}{2}+2x$

$\Rightarrow \beta=0$ for this to be circle

$(2+\alpha) \frac{x^{2}}{2}+\alpha y^{2}+2 x-4 \alpha y=0$

$\Rightarrow \alpha=2$

i.e. $2 x^{2}+2 y^{2}+2 x-8 y=0$

$x^{2}+y^{2}+x-4 y=0$

$\operatorname{radius}=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}$

Answer key and solution verified before publishing.

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