The diameters of a circle are along 2x+y-7=0 | vedclass

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(C) The intersection point of the diameter lines is the center of the circle. Solving $2x+y=7$ and $x+3y=11$: From the first equation,$y=7-2x$. Substi

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$x^2+y^2-4x-6y-12=0$

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(C) The intersection point of the diameter lines is the center of the circle. Solving $2x+y=7$ and $x+3y=11$: From the first equation,$y=7-2x$. Substituting into the second: $x+3(7-2x)=11$ $\Rightarrow x+21-6x=11$ $\Rightarrow -5x=-10$ $\Rightarrow x=2$. Then $y=7-2(2)=3$. So,the center is $(h,k) = (2,3)$. The radius $r$ is the distance between the center $(2,3)$ and the point $(5,7)$: $r = \sqrt{(5-2)^2+(7-3)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = 5$. The equation of the circle is $(x-h)^2+(y-k)^2=r^2$: $(x-2)^2+(y-3)^2 = 5^2$ $x^2-4x+4+y^2-6y+9 = 25$ $x^2+y^2-4x-6y+13-25 = 0$ $x^2+y^2-4x-6y-12 = 0$.

The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then,the equation of this circle,which also passes through $(5,7)$,is

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