If two circles x^2+y^2-6x-6y+13=0 and x^2+y^2 | vedclass

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(B) The equations of the circles are $C_1: x^2+y^2-6x-6y+13=0$ and $C_2: x^2+y^2-8y+9=0$.The equation of the common chord $AB$ is given by $C_1 - C_2

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$\left(-\frac{3}{5}, \frac{1}{5}\right)$

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(B) The equations of the circles are $C_1: x^2+y^2-6x-6y+13=0$ and $C_2: x^2+y^2-8y+9=0$.The equation of the common chord $AB$ is given by $C_1 - C_2 = 0$:$(x^2+y^2-6x-6y+13) - (x^2+y^2-8y+9) = 0$$-6x+2y+4 = 0 \Rightarrow 3x-y-2 = 0$.This is the directrix of the parabola.The slope of the directrix is $m_D = 3$.The axis of the parabola is perpendicular to the directrix and passes through the vertex $O(0,0)$.The slope of the axis is $m_A = -1/3$.The equation of the axis is $y - 0 = -1/3(x - 0) \Rightarrow x+3y = 0$.The foot of the perpendicular from the vertex to the directrix is the intersection of $3x-y-2=0$ and $x+3y=0$.Solving these,we get $x = 3/5$ and $y = -1/5$. Let this point be $Z(3/5, -1/5)$.Since the vertex $O(0,0)$ is the midpoint of the segment connecting the focus $S(\alpha, \beta)$ and the foot of the perpendicular $Z(3/5, -1/5)$:$0 = (\alpha + 3/5)/2 \Rightarrow \alpha = -3/5$$0 = (\beta - 1/5)/2 \Rightarrow \beta = 1/5$.Thus,the focus is $(-3/5, 1/5)$.

If two circles $x^2+y^2-6x-6y+13=0$ and $x^2+y^2-8y+9=0$ intersect at $A$ and $B$,then the focus of the parabola whose directrix is the line $AB$ and vertex is the point $O(0,0)$ is

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