The line $3 x+y-5=0$ touches a circle $S$ at $(1,2)$. If $(\mathrm{h}, \mathrm{k})$ is the centre of the…
The line $3 x+y-5=0$ touches a circle $S$ at $(1,2)$. If $(\mathrm{h}, \mathrm{k})$ is the centre of the circle $\mathrm{S}$ such that $\mathrm{h}^2+\mathrm{hk}+\mathrm{k}^2=$ 37 and the radius of the circle $\mathrm{S}$ is $\sqrt{10}$, then $\mathrm{k}=$
$4$
$3$
$2$
$1$
Solution
$h^2+h k+k^2=37$ ...(i)
Radius $=\sqrt{10}$
Then, equation of circle $S^{\prime}$ is
$(x-h)^2+(y-k)^2=10$ ...(ii)
$\begin{array}{ll}\because & (1,2) \text { lies on equation (ii) } \\ \therefore & (1-\mathrm{h})^2+(2-\mathrm{k})^2=10 \\ \Rightarrow & \mathrm{h}^2-2 \mathrm{~h}+1+4+\mathrm{k}^2-4 \mathrm{k}=10 \\ \Rightarrow & \mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}=9\end{array}$
$\Rightarrow \quad 37-h k-2 h-4 k=9 \quad\left\{\right.$ from eq ${ }^n(i)$
$\Rightarrow \quad h k+2 h+4 k=32$ ...(iii)
$\because$ The length of perpendicular from $(h, k)$ to $3 x+y-5=0$ is radius of the circle $S$
$\frac{3 h+k-5}{\sqrt{10}}=\sqrt{10}$
$\Rightarrow 3 \mathrm{~h}+\mathrm{k}-5=10 \Rightarrow 3 \mathrm{~h}+\mathrm{k}=1 \mathrm{~s}$ ...(iv)
Solving eq ${ }^{\mathrm{n}}$ (iv) and (iii), we get :
$\mathrm{h}=4, \mathrm{k}=3$