Question

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

• A. $\sqrt{17}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{17}}{2}$
• D. 2

Answer 1

The Correct Option is C

\[ \frac{dy}{dx} = \frac{(2 + \alpha)x - \beta y + 2}{\beta x - y(2\alpha + \beta) + 4\alpha} \]

\[ \beta xdy - (2\alpha + \beta)ydy + 4\alpha dy = (2 + \alpha)x dx - \beta ydx + 2dx \]

\[ \beta(xdy + ydx) - (2\alpha + \beta)ydy + 4\alpha dy = (2 + \alpha)x dx + 2dx \]

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

\[ \implies \beta = 0 \quad \text{for this to be a circle} \]

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

\[ \text{Coeff. of } x^2 = y^2 \implies 2 + a = 2a \implies \alpha = 2 \]

\[ \text{i.e., } 2x^2 + 2y^2 + 2x - 8y = 0 \]

\[ x^2 + y^2 + x - 4y = 0 \]

\[ \text{rd} = \sqrt{\frac{1}{4} + 4} = \frac{\sqrt{17}}{2} \]