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 $xx_1 + yy_1 - 2

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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 $xx_1 + yy_1 - 2(x + x_1) - 3(y + y_1) + 9 = 0$. Substituting $(1, 2)$: $x(1) + y(2) - 2(x + 1) - 3(y + 2) + 9 = 0$ $x + 2y - 2x - 2 - 3y - 6 + 9 = 0$ $-x - y + 1 = 0 \Rightarrow x + y - 1 = 0$. The inverse of a point $(x_1, y_1)$ is the foot of the perpendicular from the center of the circle to the polar line,but more generally,it is the point $P'$ on the line joining the center $C$ to $P$ such that $CP \cdot CP' = r^2$. Alternatively,the inverse point $(\alpha, \beta)$ is the foot of the perpendicular from the point $(1, 2)$ to the polar line $x + y - 1 = 0$. Using the formula $\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = -\frac{(1 + 2 - 1)}{1^2 + 1^2} = -\frac{2}{2} = -1$. $\alpha - 1 = -1 \Rightarrow \alpha = 0$. $\beta - 2 = -1 \Rightarrow \beta = 1$. Thus,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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