A chord AB is drawn from the point A(0,3) on | vedclass

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(B) Let the coordinates of point $M$ be $(h, k)$.Given $A = (0, 3)$ and $M = (h, k)$. Since $M$ lies on the line segment $AB$ such that $AM = 2AB$,poi

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(B) Let the coordinates of point $M$ be $(h, k)$.Given $A = (0, 3)$ and $M = (h, k)$. Since $M$ lies on the line segment $AB$ such that $AM = 2AB$,point $B$ is the midpoint of $AM$.Thus,the coordinates of $B$ are $\left( \frac{h+0}{2}, \frac{k+3}{2} \right) = \left( \frac{h}{2}, \frac{k+3}{2} \right)$.Since $B$ lies on the circle $x^2 + 4x + (y - 3)^2 = 0$,we substitute the coordinates of $B$ into the circle equation:$\left( \frac{h}{2} \right)^2 + 4\left( \frac{h}{2} \right) + \left( \frac{k+3}{2} - 3 \right)^2 = 0$$\Rightarrow \frac{h^2}{4} + 2h + \left( \frac{k-3}{2} \right)^2 = 0$$\Rightarrow \frac{h^2}{4} + 2h + \frac{(k-3)^2}{4} = 0$Multiplying by $4$,we get $h^2 + 8h + (k-3)^2 = 0$,which simplifies to $h^2 + k^2 + 8h - 6k + 9 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 8x - 6y + 9 = 0$,which represents a circle.

$A$ chord $AB$ is drawn from the point $A(0,3)$ on the circle $x^2 + 4x + (y - 3)^2 = 0$. If the point $M$ lies on the chord such that $AM = 2AB$,then the locus of point $M$ is:

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