MHT CET · Maths · Differential Equations
The differential equation of the circles having their centres on the line \(\mathrm{y}=8\) and touching the \(\mathrm{X}\) -axis is
- A \((y-8)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=64\)
- B \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
- C \((y-8)\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
- D \(y^{2}\left(1+\frac{d y}{d x}\right)=64\)
Answer & Solution
Correct Answer
(B) \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
Step-by-step Solution
Detailed explanation
(B)
Let \((\mathrm{h}, 8)\) be the centre of the circle.
Since circle touches \(X\) axis, radius \(=8\)
\(\therefore(x-h)^{2}+(y-8)^{2}=(8)^{2}\)...(1)
Differentiating w.r.t. x, we get
\(\begin{aligned}
& 2(x-h)+2(y-8) \frac{d y}{d x}=0 \\
\therefore \quad &(x-h)=-(y-8) \frac{d y}{d x}
\end{aligned}\)
Substituting value of \((x-h)\) in eq. (1), we get
\(\begin{aligned}
&\left[-(y-8) \frac{d y}{d x}\right]^{2}+(y-8)^{2}=64 \\
\therefore &(y-8)^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=64
\end{aligned}\)
Let \((\mathrm{h}, 8)\) be the centre of the circle.
Since circle touches \(X\) axis, radius \(=8\)
\(\therefore(x-h)^{2}+(y-8)^{2}=(8)^{2}\)...(1)
Differentiating w.r.t. x, we get
\(\begin{aligned}
& 2(x-h)+2(y-8) \frac{d y}{d x}=0 \\
\therefore \quad &(x-h)=-(y-8) \frac{d y}{d x}
\end{aligned}\)
Substituting value of \((x-h)\) in eq. (1), we get
\(\begin{aligned}
&\left[-(y-8) \frac{d y}{d x}\right]^{2}+(y-8)^{2}=64 \\
\therefore &(y-8)^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=64
\end{aligned}\)
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