At which point does the line x = 0 touch the | vedclass

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

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$(0, 3)$

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(C) The given equation of the circle is $x^2 + y^2 - 2x - 6y + 9 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = -3$.The center of the circle is $(-g, -f) = (1, 3)$.The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 9} = \sqrt{1 + 9 - 9} = \sqrt{1} = 1$.Since the center is $(1, 3)$ and the radius is $1$,the circle touches the $y$-axis (which is the line $x = 0$) at the point where the $x$-coordinate is $0$ and the $y$-coordinate is the same as the center's $y$-coordinate.Thus,the point of contact is $(0, 3)$.

At which point does the line $x = 0$ touch the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

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