Suppose the solution of the differential equation $\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$ represents a circle passing through origin. Then radius of the circle is

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Differential Equations

Suppose the solution of the differential equation

\frac{d y}{d x} = \frac{( 2 + α ) x - β y + 2}{β x - 2 α y - ( β γ - 4 α )}

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

Answer: (C)

Writing the differential equation in exact form and integrating (choosing the constant so the solution family is a circle through the origin) gives

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

. For a circle

x^{2} + y^{2} + D x + E y + F = 0

, the radius is

(D /2)^{2} + (E /2)^{2} - F = (\frac{1}{2})^{2} + 2^{2} = \frac{17}{4} = \frac{17}{2}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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