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 P, such that AP=2AB. The locus of P is

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing clear explanations for each step.

### Step 1: Identify the Circle Equation
The given equation of the circle is:
\[ x^2 + 4x + (y - 3)^2 = 0 \]

### Step 2: Rewrite the Circle Equation
We can rewrite the equation in a more recognizable form by completing the square for the x-terms:
\[ x^2 + 4x = (x + 2)^2 - 4 \]
Thus, the equation becomes:
\[ (x + 2)^2 - 4 + (y - 3)^2 = 0 \]
This simplifies to:
\[ (x + 2)^2 + (y - 3)^2 = 4 \]
This represents a circle with center at (-2, 3) and radius 2.

### Step 3: Identify Point A
The point A is given as \( A(0, 3) \), which lies on the circle.

### Step 4: Define Point B
Let point B be represented as \( B(h, k) \). Since AP = 2AB, point B is the midpoint of segment AP. Therefore, we can express the coordinates of B as:
\[ B\left(\frac{0 + h}{2}, \frac{3 + k}{2}\right) = \left(\frac{h}{2}, \frac{k + 3}{2}\right) \]

### Step 5: Substitute B into the Circle Equation
Since point B lies on the circle, we substitute \( x = \frac{h}{2} \) and \( y = \frac{k + 3}{2} \) into the circle's equation:
\[ \left(\frac{h}{2}\right)^2 + 4\left(\frac{h}{2}\right) + \left(\frac{k + 3}{2} - 3\right)^2 = 0 \]

### Step 6: Simplify the Equation
This leads to:
\[ \frac{h^2}{4} + 2h + \left(\frac{k - 3}{2}\right)^2 = 0 \]
Expanding the square:
\[ \frac{h^2}{4} + 2h + \frac{(k - 3)^2}{4} = 0 \]

### Step 7: Clear the Denominator
Multiply the entire equation by 4 to eliminate the fractions:
\[ h^2 + 8h + (k - 3)^2 = 0 \]

### Step 8: Rearranging the Equation
Rearranging gives:
\[ h^2 + (k - 3)^2 + 8h = 0 \]

### Step 9: Completing the Square
Completing the square for the \( h \) terms:
\[ (h + 4)^2 - 16 + (k - 3)^2 = 0 \]
This can be rewritten as:
\[ (h + 4)^2 + (k - 3)^2 = 16 \]

### Step 10: Identify the Locus of Point P
The equation \( (h + 4)^2 + (k - 3)^2 = 16 \) represents a circle centered at (-4, 3) with a radius of 4.

### Final Answer
The locus of point P is:
\[ (x + 4)^2 + (y - 3)^2 = 16 \]