The radius of the circle which passes through the focus of parabola ${\mathrm{x}}^2=4\mathrm{y}$ and touches it at point (6,9) is $\mathrm{k}\sqrt{10}$ then k =

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{x}}^2=4\mathrm{y} \\ \mathrm{Focus}=(0,1) \end{gathered}$$

$$\begin{gathered} \mathrm{Eq}uation \mathrm{of} \mathrm{tangent} \mathrm{at} P \mathrm{is} S_1=0 \\ \Rightarrow 6\mathrm{x}-2\left(\mathrm{y}+9\right)=0 \\ \Rightarrow 3\mathrm{x}-\mathrm{y}-9=0 \\ The equation of circle touching at (6,9) is \end{gathered}$$

${\left(\mathrm{x}-6\right)}^2+{\left(\mathrm{y}-9\right)}^2+\mathrm{\lambda }\left(3\mathrm{x}-\mathrm{y}-9\right)=0$

it is passing through $(0,1) then 36+64+\mathrm{\lambda }\left(-10\right)=0$

$$\begin{gathered} \Rightarrow \mathrm{\lambda }=10. \\ then circle is x^2+y^2+18x-28y+27=0 \\ radius \mathrm{r}=\sqrt{250}=5\sqrt{10} \\ BY comparision \mathrm{k}=5 \end{gathered}$$