If the order and degree of the differential equation corresponding to the family of curves $(x-2)^2+(y-a)^2=b^2$, (where $a$ and $b$ are parameters) are $m$ and $n$ respectively, then $m^2+n=$

• A

  7
• Correct answer
  B

  5
• C

  4
• D

  3

Solution

Given, $(x-2)^2+(y-a)^2=b^2$ On differentiating w.r.t. ‘x’
Again differentiating w.r.t ‘x’
From Eqs. (i) and (ii) $\begin{aligned} & \quad 1+\frac{-(x-2) y^{\prime \prime}}{y^{\prime}}+y^{\prime 2}=0 \Rightarrow \\ & (x-2) y^{\prime \prime}=y^{\prime 3}+y^{\prime} \\ & \because \text { Order }=2, \text { degree }=1 \\ & \therefore \quad m^2+n=(2)^2+1=5 \end{aligned}$

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