If Q(h, k) is the inverse point of the point | vedclass

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(B) The given equation of the circle is $x^2+y^2-4x+1=0$. Rewriting it in the form $(x-g)^2+(y-f)^2=r^2$,we get $(x-2)^2+y^2=3$. Thus,the center $C =

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$4$

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(B) The given equation of the circle is $x^2+y^2-4x+1=0$. Rewriting it in the form $(x-g)^2+(y-f)^2=r^2$,we get $(x-2)^2+y^2=3$. Thus,the center $C = (2, 0)$ and the radius squared $r^2 = 3$. The inverse point $Q(h, k)$ of $P(x_1, y_1)$ with respect to a circle with center $(x_0, y_0)$ and radius $r$ is given by the formula: $h = x_0 + \frac{r^2(x_1-x_0)}{(x_1-x_0)^2+(y_1-y_0)^2}$ and $k = y_0 + \frac{r^2(y_1-y_0)}{(x_1-x_0)^2+(y_1-y_0)^2}$. Here,$(x_0, y_0) = (2, 0)$,$(x_1, y_1) = (1, 2)$,and $r^2 = 3$. The denominator is $(1-2)^2+(2-0)^2 = (-1)^2 + 2^2 = 1+4 = 5$. So,$h = 2 + \frac{3(1-2)}{5} = 2 - \frac{3}{5} = \frac{7}{5}$. And $k = 0 + \frac{3(2-0)}{5} = \frac{6}{5}$. Finally,$2h+k = 2(\frac{7}{5}) + \frac{6}{5} = \frac{14}{5} + \frac{6}{5} = \frac{20}{5} = 4$.

If $Q(h, k)$ is the inverse point of the point $P(1, 2)$ with respect to the circle $x^2+y^2-4x+1=0$,then $2h+k=$

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