Q63 / 105 · JEE Main 2008 · Mathematics

The differential equation of the family of circles with fixed radius 5 units and centre on the line $y=2$ is

  • A
    $(x-2) y^{\prime 2}=25-(y-2)^2$
  • B
    $(y-2) y^{\prime 2}=25-(y-2)^2$
  • Correct answer
    C
    $(y-2)^2 y^{\prime 2}=25-(y-2)^2$
  • D
    $(x-2)^2 y^{\prime 2}=25-(y-2)^2$

Solution

$$ \begin{aligned} & (x-h)^2+(y-2)^2=25 \\ & \Rightarrow 2(x-h)+2(y-2) \frac{d y}{d x}=0 \\ & \Rightarrow(x-h)=-(y-2) \frac{d y}{d x} \end{aligned} $$ substituting in (1), we have $$ \begin{aligned} & (y-2)^2\left(\frac{d y}{d x}\right)^2+(y-2)^2=25 \\ & (y-2)^2 y^{\prime 2}=25-(y-2)^2 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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