Let the coordinates of the centre of the circle be $\left(h,k\right)$ and radius be $r$. Then, equation of circle is

${\left(x-h\right)}^2+{\left(y-k\right)}^2=5^2 ..\left(1\right)$

Differentiating both sides w.r.t. $x$, we get

$2\left(x-h\right)+2\left(y-k\right)y^{\text{'}}=0$

$\Rightarrow \left(x-h\right)=-\left(y-k\right)y^{\text{'}} ...\left(i\right)$

Again differentiating w.r.t. $x$, we get

$1=-\left\{{\left(y^{\text{'}}\right)}^2+\left(y-k\right)y^{\text{'}\text{'}}\right\}$

$\Rightarrow \left(y-k\right)=-\left(\frac{1+{\left(y^{\text{'}}\right)}^2}{y^{\text{'}\text{'}}}\right) ...\left(ii\right)$

Now, equation $\left(i\right)$ becomes 

$\Rightarrow \left(x-h\right)=\left(\frac{1+{\left(y^{\text{'}}\right)}^2}{y^{\text{'}\text{'}}}\right)y^{\text{'}} ...\left(iii\right)$

Substituting the values of $\left(x-h\right) \& \left(y-k\right)$ from $\left(ii\right) \& \left(iii\right)$ in $\left(1\right)$, we get

${\left[\left(\frac{1+{\left(y^{\text{'}}\right)}^2}{y^{\text{'}\text{'}}}\right)y^{\text{'}}\right]}^2+{\left[-\left(\frac{1+{\left(y^{\text{'}}\right)}^2}{y^{\text{'}\text{'}}}\right)\right]}^2=25$

$\Rightarrow \frac{{\left\{1+{\left(y^{\text{'}}\right)}^2\right\}}^3}{{\left(y^{\text{'}\text{'}}\right)}^2}=25$

$\Rightarrow {\left\{1+{\left(y^{\text{'}}\right)}^2\right\}}^3=25{\left(y^{\text{'}\text{'}}\right)}^2$

Highest order derivative is of order $2$. Therefore, order of differential equation is $2$. Hence,

$m=2$

Highest order derivative is of order $2$ whose degree is $2$. Therefore, degree of differential equation is $2$. Thus,

$l=2$

And,

$2l+3m=10$