MHT CET · Maths · Differential Equations
The differential equation of all circles, passing through the origin and having their centres on the X-axis, is
- A \(y^2=x^2+x y \frac{\mathrm{d} y}{\mathrm{~d} x}\)
- B \(x^2=y^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}\)
- C \(y^2=x^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}\)
- D \(x^2=y^2-x y \frac{\mathrm{d} y}{\mathrm{~d} x}\)
Answer & Solution
Correct Answer
(C) \(y^2=x^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}\)
Step-by-step Solution
Detailed explanation
The system of circles which passes through origin and whose centre lies on \(\mathrm{X}\)-axis is \(x^2+y^2-2 b x=0... (i)\)
Differentiating w.r.t \(x\), we get
\(2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{~b}... (ii)\)
Substituting (ii) in (i), we get
\(\begin{aligned}
& x^2+y^2-2 x^2-2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \Rightarrow y^2-x^2-2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \Rightarrow y^2=x^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}
\end{aligned}\)
Differentiating w.r.t \(x\), we get
\(2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{~b}... (ii)\)
Substituting (ii) in (i), we get
\(\begin{aligned}
& x^2+y^2-2 x^2-2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \Rightarrow y^2-x^2-2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \Rightarrow y^2=x^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}
\end{aligned}\)
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