If $\mathrm{Q}(h, k)$ is the inverse point of the point $\mathrm{P}(1,2)$ with respect to the circle…
If $\mathrm{Q}(h, k)$ is the inverse point of the point $\mathrm{P}(1,2)$ with respect to the circle $x^2+y^2-4 x+1-0$, then $2 h+k=$
$3$
$4$
$7$
$11$
Solution
Given equation of circle $x^2+y^2-4 x+1=0$ $\Rightarrow(x-2)^2+y^2=3$. So, centre $=(2,0)$, radius $=\sqrt{3}$
Now, inverse point is
$\mathrm{h}=\alpha(1-2)+2 \Rightarrow k=\alpha(2-0)+0$
where, $\alpha=\frac{r^2}{(1-2)^2+(2-0)^2}=\frac{3}{1+4}=\frac{3}{5}$
Now, $h=\frac{3}{5}(1-2)+2=\frac{-3}{5}+2=\frac{7}{5}$ and $k=\frac{3}{5}(2)=\frac{6}{5}$
So, $2 h+k=\frac{14}{5}+\frac{6}{5}=\frac{20}{5}=4$