The inverse of the point (1, 2) with respect | vedclass

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(C) The equation of the polar of the point $(x_1, y_1) = (1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is given by $T = 0$: $x(1) +

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$(0, 1)$

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(C) The equation of the polar of the point $(x_1, y_1) = (1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is given by $T = 0$: $x(1) + y(2) - 2(x + 1) - 3(y + 2) + 9 = 0$ $x + 2y - 2x - 2 - 3y - 6 + 9 = 0$ $-x - y + 1 = 0 \Rightarrow x + y - 1 = 0$. The inverse point $(\alpha, \beta)$ is the foot of the perpendicular from the point $(1, 2)$ to the line $x + y - 1 = 0$. Using the formula $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$: $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = -\frac{1(1) + 1(2) - 1}{1^2 + 1^2}$ $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = -\frac{2}{2} = -1$. Thus,$\alpha - 1 = -1 \Rightarrow \alpha = 0$ and $\beta - 2 = -1 \Rightarrow \beta = 1$. The inverse point is $(0, 1)$.

The inverse of the point $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

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