TS EAMCET · Maths · Differential Equations
The differential equation corresponding to all the circles lying in the first quadrant and touching the coordinate axes is
- A \((x-y)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(x+y \frac{d y}{d x}\right)^2\)
- B \((x-y)^2\left[1+\frac{d y}{d x}\right]^2=\left(x+y \frac{d y}{d x}\right)^2\)
- C \((x-y)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=x+y\left(\frac{d y}{d x}\right)^2\)
- D \((x-y)^2\left[1+\frac{d y}{d x}\right]=\left(x+y \frac{d y}{d x}\right)^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(A) \((x-y)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(x+y \frac{d y}{d x}\right)^2\)
Step-by-step Solution
Detailed explanation
Let the radius at circle \(a\), then centre of the circle is \((a, a)\). Hence, the equation of the circle is \( \begin{aligned} (x-a)^2+(y-a)^2 & =a^2 \\ \Rightarrow \quad x^2+y^2-2 a x-2 a y+a^2 & =0 \end{aligned} \) On differentiating Eq. (i) w.r.t. \(x\), we get…
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