JEE Advanced · Mathematics · 29. Differential Eqns
The differential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}\) determines a family of circles with
- A
variable radii and a fixed centre at \((0,1)\)
- B
variable radii and a fixed centre at \((0,-1)\)
- C
fixed radius 1 and variable centres along the \(\mathrm{X}\)-axis
- D
fixed radius 1 and variable centres along the \(\mathrm{Y}\)-axis
Answer & Solution
Correct Answer
(C)
fixed radius 1 and variable centres along the \(\mathrm{X}\)-axis
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
& \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} \\
& \Rightarrow \quad \int \frac{y}{\sqrt{1-y^2}} d y=\int d x \\
& \Rightarrow \quad-\sqrt{1-y^2}=x+C \\
& \Rightarrow \quad(x+c)^2+y^2=1 \\
&
\end{aligned}
\]
Centre \((-C, 0)\) and radius \(=1\).
\begin{aligned}
& \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} \\
& \Rightarrow \quad \int \frac{y}{\sqrt{1-y^2}} d y=\int d x \\
& \Rightarrow \quad-\sqrt{1-y^2}=x+C \\
& \Rightarrow \quad(x+c)^2+y^2=1 \\
&
\end{aligned}
\]
Centre \((-C, 0)\) and radius \(=1\).
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