JEE Mains · Maths · STD 12 - 9. differential equations
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 :
- A \(\sqrt{17}\)
- B \(\frac{1}{2}\)
- C \(\frac{\sqrt{17}}{2}\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{17}}{2}\)
Step-by-step Solution
Detailed explanation
\( \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-y(2 \alpha+\beta)+4 \alpha} \) \( \beta x d y-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x-\beta y d x+2 d x \) \( \beta(x d y+y d x)-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x+2 d x \)…
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