If the segments of the straight lines x+y=6 a | vedclass

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(C) Let $L_1 \equiv x+y=6$ and $L_2 \equiv x+2y=4$.The point of intersection of the two diameters $L_1$ and $L_2$ is the center $C$ of the circle.Solv

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$x^2+y^2-16x+4y+48=0$

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(C) Let $L_1 \equiv x+y=6$ and $L_2 \equiv x+2y=4$.The point of intersection of the two diameters $L_1$ and $L_2$ is the center $C$ of the circle.Solving the system of equations:$x+y=6 \Rightarrow x=6-y$Substituting into $L_2$: $(6-y)+2y=4 \Rightarrow y=-2$.Then $x=6-(-2)=8$.Thus,the center $C$ is $(8, -2)$.The circle passes through the point $P(6, 2)$. The radius $r$ is the distance $CP$.$r^2 = CP^2 = (8-6)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$.The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.$(x-8)^2 + (y+2)^2 = 20$$x^2 - 16x + 64 + y^2 + 4y + 4 = 20$$x^2 + y^2 - 16x + 4y + 48 = 0$.

If the segments of the straight lines $x+y=6$ and $x+2y=4$ are two diameters of a circle passing through $(6,2)$,then the equation of that circle is

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