Question

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $X Y$ plane with radius 5 units, then $2 l+3 m=$

• A. 5
• B. 10
• C. 15
• D. 7

Answer 1

Answer: (B)

Family of all circles in the $x y$ -planewith radius $5$ units with center $\left(x_{1}, y_{1}\right)$ is
$\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}=5^{2} $
$\Rightarrow \left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}=25\,\,\,...(i)$
differential w.r.t ' $x^{'}$
$\Rightarrow 2\left(x-x_{1}\right)+2\left(y-y_{1}\right) y^{'}=0 \,\,\,\cdots(ii)$
$\Rightarrow \left(x-x_{1}\right)=-\left(y-y_{1}\right) y^{'}$
Now Eq. (ii) diff again w.r.t ' $x^{'}$,
we get
$2+2\left(y^{''}\left(y-y_{1}\right)+\left(y^{'}\right)^{2}\right] =0 $
$\Rightarrow \left(y-y_{1}\right) =\frac{1-\left(y^{'}\right)^{2}}{y^{ ''}}$
Sub. values in Eq. (i)
$=\left[\frac{1+\left(y^{'}\right)^{2}}{y^{''}} \cdot y^{'}\right]^{2}+\left[\frac{1+\left(y^{'}\right)^{2}}{y^{''}}\right]=25$
$\Rightarrow \frac{\left(1+y^{'2}\right)}{\left(y^{''}\right)^{2}} \cdot\left(y^{'}\right)^{2}+\frac{1+\left(y^{'}\right)^{2}}{\left(y^{''}\right)^{2}}=25$
$\Rightarrow \frac{\left(1+y^{'}\right)^{3}}{\left(y^{''}\right)^{2}}=25$
$\Rightarrow 25\left(y^{''}\right)^{2}=\left[1+\left(y^{'}\right)^{2}\right]^{3}$
So, order $=2$ and degree $=2$
$\therefore l=2$ and $m=2$
Now, $2 l+3 m=2 \times 2+3 \times 2$
$=4+6=10 $