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(D) The equation of a family of circles passing through the intersection of the circle $S \equiv x^2+y^2-2x-6y+3=0$ and the line $L \equiv 2x+y=0$ is

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$(-2, 1)$

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(D) The equation of a family of circles passing through the intersection of the circle $S \equiv x^2+y^2-2x-6y+3=0$ and the line $L \equiv 2x+y=0$ is given by $S + \lambda L = 0$. $x^2+y^2-2x-6y+3 + \lambda(2x+y) = 0$ $x^2+y^2 + x(2\lambda-2) + y(\lambda-6) + 3 = 0$. Since the chord $2x+y=0$ is the diameter of this new circle,the center of the circle must lie on the line $2x+y=0$. The center of the circle is $(1-\lambda, \frac{6-\lambda}{2})$. Substituting the center into the line equation: $2(1-\lambda) + \frac{6-\lambda}{2} = 0$. $4 - 4\lambda + 6 - \lambda = 0$ $\Rightarrow 5\lambda = 10$ $\Rightarrow \lambda = 2$. Substituting $\lambda = 2$ into the circle equation: $x^2+y^2 + x(4-2) + y(2-6) + 3 = 0$. $x^2+y^2+2x-4y+3 = 0$. Checking the options,for point $(-2, 1)$: $(-2)^2 + (1)^2 + 2(-2) - 4(1) + 3 = 4 + 1 - 4 - 4 + 3 = 0$. Thus,the circle passes through the point $(-2, 1)$.

If $2x+y=0$ is the equation of a chord of the circle $x^2+y^2-2x-6y+3=0$,then the circle with this chord as diameter passes through the point

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