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

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(C) Let the coordinates of $M$ be $(x, y)$.Given $A(0, 3)$ and $AM = 2AB$, this implies that $B$ is the midpoint of the segment $AM$.Therefore, the co

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$x^{2}+y^{2}+8x-6y+9=0$

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(C) Let the coordinates of $M$ be $(x, y)$.Given $A(0, 3)$ and $AM = 2AB$, this implies that $B$ is the midpoint of the segment $AM$.Therefore, the coordinates of $B$ are $\left(\frac{0+x}{2}, \frac{3+y}{2}\right) = \left(\frac{x}{2}, \frac{y+3}{2}\right)$.Since $B$ lies on the circle $x^{2}+4x+(y-3)^{2}=0$, we substitute the coordinates of $B$ into the circle's equation:$\left(\frac{x}{2}\right)^{2} + 4\left(\frac{x}{2}\right) + \left(\frac{y+3}{2} - 3\right)^{2} = 0$$\Rightarrow \frac{x^{2}}{4} + 2x + \left(\frac{y+3-6}{2}\right)^{2} = 0$$\Rightarrow \frac{x^{2}}{4} + 2x + \left(\frac{y-3}{2}\right)^{2} = 0$$\Rightarrow \frac{x^{2}}{4} + 2x + \frac{y^{2}-6y+9}{4} = 0$Multiplying the entire equation by $4$, we get:$x^{2} + 8x + y^{2} - 6y + 9 = 0$Thus, the locus of $M$ is $x^{2}+y^{2}+8x-6y+9=0$.

$A$ chord $AB$ is drawn from the point $A(0,3)$ on the circle $x^{2}+4x+(y-3)^{2}=0$, and is extended to $M$ such that $AM=2AB$. The locus of $M$ is

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