If the inverse point of (1,1) with respect to | vedclass

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(A) The equation of the polar of the point $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is given by $x(1) + y(1) - 2(x+1) - 3(y+1) + 12 = 0

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$\frac{22}{5}$

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(A) The equation of the polar of the point $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is given by $x(1) + y(1) - 2(x+1) - 3(y+1) + 12 = 0$.Simplifying this,we get $x + y - 2x - 2 - 3y - 3 + 12 = 0$,which results in $-x - 2y + 7 = 0$,or $x + 2y - 7 = 0$.The inverse point $(h, k)$ is the foot of the perpendicular from the point $(1,1)$ to the polar line $x + 2y - 7 = 0$.Using the formula for the foot of the perpendicular $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$:$\frac{h-1}{1} = \frac{k-1}{2} = -\frac{1(1) + 2(1) - 7}{1^2 + 2^2} = -\frac{1 + 2 - 7}{5} = -\frac{-4}{5} = \frac{4}{5}$.From $\frac{h-1}{1} = \frac{4}{5}$,we get $h = 1 + \frac{4}{5} = \frac{9}{5}$.From $\frac{k-1}{2} = \frac{4}{5}$,we get $k - 1 = \frac{8}{5}$,so $k = 1 + \frac{8}{5} = \frac{13}{5}$.Therefore,$h + k = \frac{9}{5} + \frac{13}{5} = \frac{22}{5}$.

If the inverse point of $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is $(h, k)$,then $h+k$ is equal to

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