MHT CET · Maths · Differential Equations
The differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y}\) determines a family of circles with
- A variable radii and fixed centre at \((0,1)\).
- B variable radii and fixed centre at \((0,-1)\).
- C fixed radius of 1 unit and variable centre along the \(\mathrm{Y}\)-axis.
- D fixed radius of 1 unit and variable centre along the \(\mathrm{X}\)-axis.
Answer & Solution
Correct Answer
(D) fixed radius of 1 unit and variable centre along the \(\mathrm{X}\)-axis.
Step-by-step Solution
Detailed explanation
\( \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y} \)
\( \therefore \int \frac{y}{\sqrt{1-y^2}} \mathrm{~d} y=\int 1 \mathrm{~d} x \)
\( \therefore -\sqrt{1-y^2}=x+\mathrm{c} \)
\( \therefore (x+\mathrm{c})^2=1-y^2\)
\(\therefore (x+\mathrm{c})^2+y^2=1\)
\(\therefore\) Radius is fixed, which is 1 and the centre is \((-c, 0)\) which is a variable centre on the \(\mathrm{X}\)-axis.
\( \therefore \int \frac{y}{\sqrt{1-y^2}} \mathrm{~d} y=\int 1 \mathrm{~d} x \)
\( \therefore -\sqrt{1-y^2}=x+\mathrm{c} \)
\( \therefore (x+\mathrm{c})^2=1-y^2\)
\(\therefore (x+\mathrm{c})^2+y^2=1\)
\(\therefore\) Radius is fixed, which is 1 and the centre is \((-c, 0)\) which is a variable centre on the \(\mathrm{X}\)-axis.
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