Suppose the solution of the differential equation represents a circle passing through origin. Then the radius of this circle is: [2024]

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Solution

Q.
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 the radius of this circle is: [2024]

1 2

2 $\frac{1}{2}$

3 $\sqrt{17}$

4 $\frac{\sqrt{17}}{2}$

Ans.
(4)

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

$\Rightarrow β x d y - 2 α y d y - (β γ - 4 α) d y = 2 x d x + α x d x - β y d x + 2 d x$

On integrating, we get

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

$\Rightarrow β x y - α y^{2} - (β γ - 4 α) y = x^{2} (1 + \frac{α}{2}) + 2 x + c$

$\Rightarrow (1 + \frac{α}{2}) x^{2} + α y^{2} - β x y + 2 x + (β γ - 4 α) y + c = 0$

Since, it represents a circle which is passing through origin, then

$1 + \frac{α}{2} = α, β = 0 and c = 0 [\begin{matrix} ∵ coeff . of x^{2} = coeff . of y^{2} \\ coeff . of x y = 0 \end{matrix}]$

$\Rightarrow α = 2$

$∴$ Equation of circle is given by

$2 x^{2} + 2 y^{2} + 2 x - 8 y = 0 \Rightarrow x^{2} + y^{2} + x - 4 y = 0$

$\Rightarrow Centre \equiv (- \frac{1}{2}, 2)$

$\Rightarrow Radius = \sqrt{{(\frac{1}{2})}^{2} + {(2)}^{2} - 0} = \frac{\sqrt{17}}{2}$

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