What is the distance in the xy-plane from the point `(3, -6)` to the center of the circle whose equation is `x(x+4)+y(y-12)=9`?

Answer

To find the distance in the xy-plane from the point (3, -6) to the center of the circle given by the equation \( x(x+4) + y(y-12) = 9 \), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
The given equation of the circle is:
\[ x(x + 4) + y(y - 12) = 9 \]
This can be expanded and rearranged as:
\[ x^2 + 4x + y^2 - 12y = 9 \]

### Step 2: Complete the Square for x and y
To convert this into the standard form of a circle, we will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + 4x \rightarrow (x + 2)^2 - 4 \]

For \(y\):
\[ y^2 - 12y \rightarrow (y - 6)^2 - 36 \]

### Step 3: Substitute Back into the Equation
Now substitute these completed squares back into the equation:
\[ (x + 2)^2 - 4 + (y - 6)^2 - 36 = 9 \]
Combine like terms:
\[ (x + 2)^2 + (y - 6)^2 - 40 = 9 \]
Add 40 to both sides:
\[ (x + 2)^2 + (y - 6)^2 = 49 \]

### Step 4: Identify the Center of the Circle
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify the center of the circle:
- \(h = -2\)
- \(k = 6\)

Thus, the center of the circle is at the point \((-2, 6)\).

### Step 5: Use the Distance Formula
Now we will find the distance \(d\) from the point \((3, -6)\) to the center \((-2, 6)\) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, \((x_1, y_1) = (3, -6)\) and \((x_2, y_2) = (-2, 6)\).

Substituting the values:
\[ d = \sqrt{((-2) - 3)^2 + (6 - (-6))^2} \]
\[ = \sqrt{(-5)^2 + (12)^2} \]
\[ = \sqrt{25 + 144} \]
\[ = \sqrt{169} \]
\[ = 13 \]

### Final Answer
The distance from the point \((3, -6)\) to the center of the circle is \(13\).

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