From the point A(0, 3) on the circle x^2 + 4x | vedclass

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(D) The given circle equation is $x^2 + 4x + (y - 3)^2 = 0$. Let $M = (h, k)$ and $A = (0, 3)$. Since $AM = 2 AB$,$B$ is the midpoint of $AM$. Thus,th

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

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(D) The given circle equation is $x^2 + 4x + (y - 3)^2 = 0$. Let $M = (h, k)$ and $A = (0, 3)$. Since $AM = 2 AB$,$B$ is the midpoint of $AM$. Thus,the coordinates of $B$ are $(\frac{h+0}{2}, \frac{k+3}{2}) = (\frac{h}{2}, \frac{k+3}{2})$. Since $B$ lies on the circle,we substitute these coordinates into the circle equation: $(\frac{h}{2})^2 + 4(\frac{h}{2}) + (\frac{k+3}{2} - 3)^2 = 0$. $\frac{h^2}{4} + 2h + (\frac{k-3}{2})^2 = 0$. $\frac{h^2}{4} + 2h + \frac{k^2 - 6k + 9}{4} = 0$. Multiplying by $4$,we get $h^2 + 8h + k^2 - 6k + 9 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 + 8x - 6y + 9 = 0$.

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

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