JEE Mains · Maths · STD 12 - 9. differential equations
The differential equation of the family of circles passing through the points \((0,2)\) and \((0,-2)\) is
- A \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
- B \(2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0\)
- C \(2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0\)
- D \(2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0\)
Answer & Solution
Correct Answer
(A) \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
Step-by-step Solution
Detailed explanation
Equation of circle passing through \((0,-2)\) and \((0,2)\) is \(x ^{2}+\left( y ^{2}-4\right)+\lambda x =0,(\lambda \in R )\) Divided by \(x\) we get \(\frac{ x ^{2}+\left( y ^{2}-4\right)}{ x }+\lambda=0\) Differentiating with respect to \(x\)…
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