KCET · Maths · Differential Equations
The differential equation of the family of circles passing through the orign and having their centres on the \(x\)-axis is
- A \(\mathrm{y}^{2}=\mathrm{x}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}\)
- B \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\)
- C \(x^{2}=y^{2}+x y \frac{d y}{d x}\)
- D \(x^{2}=y^{2}+3 x y \frac{d y}{d x}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{y}^{2}=\mathrm{x}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}\)
Step-by-step Solution
Detailed explanation
The system of circles passing through origin and centre lies on \(\mathrm{x}\)-axis is
\[
\mathrm{x}^{2}+\mathrm{y}^{2}-2 h \mathrm{x}=0
\quad \text{...(i)}
\]
On differentiating w.r.t. \(x\), we get
\[
\begin{aligned}
&2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{~h}=0 \\
&\Rightarrow 2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-\left(\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{\mathrm{x}}\right)=0 \quad \text { [from Eq. (i)] } \\
&\Rightarrow \quad \mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0
\end{aligned}
\]
\[
\mathrm{x}^{2}+\mathrm{y}^{2}-2 h \mathrm{x}=0
\quad \text{...(i)}
\]
On differentiating w.r.t. \(x\), we get
\[
\begin{aligned}
&2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{~h}=0 \\
&\Rightarrow 2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-\left(\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{\mathrm{x}}\right)=0 \quad \text { [from Eq. (i)] } \\
&\Rightarrow \quad \mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=0
\end{aligned}
\]
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