The inverse point of (1, 3) with respect to t | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 4x - 6y + 9 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -3$,and $c =

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$(-2, 3)$

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(C) The equation of the circle is $x^2 + y^2 - 4x - 6y + 9 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -3$,and $c = 9$. The center of the circle is $C(-g, -f) = (2, 3)$ and the radius is $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - 9} = \sqrt{4 + 9 - 9} = 2$. Let $P(x_1, y_1) = (1, 3)$ be the given point. The inverse point $P'(x', y')$ of $P$ with respect to the circle is given by the formula: $x' - h = \frac{r^2(x_1 - h)}{(x_1 - h)^2 + (y_1 - k)^2}$ and $y' - k = \frac{r^2(y_1 - k)}{(x_1 - h)^2 + (y_1 - k)^2}$,where $(h, k)$ is the center $(2, 3)$. Substituting the values: $x' - 2 = \frac{2^2(1 - 2)}{(1 - 2)^2 + (3 - 3)^2} = \frac{4(-1)}{(-1)^2 + 0} = -4 \implies x' = -2$. $y' - 3 = \frac{2^2(3 - 3)}{(1 - 2)^2 + (3 - 3)^2} = \frac{4(0)}{1} = 0 \implies y' = 3$. Thus,the inverse point is $(-2, 3)$.

The inverse point of $(1, 3)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

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