The line 4x - 3y + 2 = 0 intersects the circl | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 2x + 6y + c = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c' = 0$,we get $g = -1$,$f = 3$,and center $O

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$2$

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(C) The equation of the circle is $x^2 + y^2 - 2x + 6y + c = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c' = 0$,we get $g = -1$,$f = 3$,and center $O = (1, -3)$. The radius $r$ is given by $r = \sqrt{g^2 + f^2 - c'} = \sqrt{1 + 9 - c} = \sqrt{10 - c}$. The perpendicular distance $d$ from the center $(1, -3)$ to the line $4x - 3y + 2 = 0$ is $d = \frac{|4(1) - 3(-3) + 2|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 + 9 + 2|}{5} = \frac{15}{5} = 3$. In a circle,$r^2 = d^2 + (AB/2)^2$. Given $AB = 8$,so $AB/2 = 4$. $r^2 = 3^2 + 4^2 = 9 + 16 = 25$. Thus,$10 - c = 25$,which gives $c = -15$. The circle equation is $x^2 + y^2 - 2x + 6y - 15 = 0$. Since $(1, k)$ lies on the circle,substitute $x = 1$ and $y = k$: $1^2 + k^2 - 2(1) + 6k - 15 = 0$ $1 + k^2 - 2 + 6k - 15 = 0$ $k^2 + 6k - 16 = 0$ $(k + 8)(k - 2) = 0$. Since $k > 0$,we have $k = 2$.

The line $4x - 3y + 2 = 0$ intersects the circle $x^2 + y^2 - 2x + 6y + c = 0$ at two points $A$ and $B$,and the length of the chord $AB = 8$. If $(1, k)$ is a point on the given circle and $k > 0$,then $k =$

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