The line `3x + 5y +9=0` w.rt. the circle `x^(2)+y^(2)-4x+6y+5=0` is

Options

A. chord
B. diameter
C. tangent
D. none

Answer

To determine the relationship between the line \(3x + 5y + 9 = 0\) and the circle defined by the equation \(x^2 + y^2 - 4x + 6y + 5 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the circle equation:
\[ x^2 + y^2 - 4x + 6y + 5 = 0 \]
We can rewrite this in the standard form by completing the square for both \(x\) and \(y\).

1. For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

2. For \(y^2 + 6y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 + 5 = 0 \]
Simplifying this, we have:
\[ (x - 2)^2 + (y + 3)^2 - 8 = 0 \implies (x - 2)^2 + (y + 3)^2 = 8 \]

### Step 2: Identify the Center and Radius
From the standard form of the circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (2, -3)\)
- Radius \(r = \sqrt{8} = 2\sqrt{2}\)

### Step 3: Find the Distance from the Center to the Line
Now, we need to find the distance from the center of the circle \((2, -3)\) to the line \(3x + 5y + 9 = 0\). The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 3\), \(B = 5\), \(C = 9\), and the point is \((2, -3)\).

Calculating the distance:
\[ d = \frac{|3(2) + 5(-3) + 9|}{\sqrt{3^2 + 5^2}} = \frac{|6 - 15 + 9|}{\sqrt{9 + 25}} = \frac{|0|}{\sqrt{34}} = 0 \]

### Step 4: Conclusion
Since the distance \(d = 0\), this means that the line passes through the center of the circle. Therefore, the line \(3x + 5y + 9 = 0\) acts as a diameter of the circle.

### Final Answer
The line \(3x + 5y + 9 = 0\) is a diameter of the circle \(x^2 + y^2 - 4x + 6y + 5 = 0\).

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