Let the equation of circle is `x^(2) + y^(2) - 6x - 4y + 9 = 0` . Then the line 4x + 3y - 8 = 0 is a

Options

A. tangent of the circle
B. normal of the circle
C. chord of the angle
D. None of the above

Explanation

To solve the problem, we need to determine the relationship between the given circle and the line. We will follow these steps:

### Step 1: Write the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 6x - 4y + 9 = 0 \]

We can rearrange it to find the center and radius. The standard form of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

### Step 2: Complete the square for the circle's equation
Rearranging the equation:
\[ x^2 - 6x + y^2 - 4y + 9 = 0 \]

Completing the square for \(x\) and \(y\):
1. For \(x^2 - 6x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
2. For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting back into the equation:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 + 9 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 2)^2 - 4 = 0 \]
Thus, we have:
\[ (x - 3)^2 + (y - 2)^2 = 4 \]

### Step 3: Identify the center and radius of the circle
From the equation \((x - 3)^2 + (y - 2)^2 = 4\):
- The center \((h, k)\) is \((3, 2)\)
- The radius \(r\) is \(\sqrt{4} = 2\)

### Step 4: Find the distance from the center of the circle to the line
The equation of the line is:
\[ 4x + 3y - 8 = 0 \]

We can use the formula for the distance \(D\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Here, \(A = 4\), \(B = 3\), \(C = -8\), and the center of the circle is \((x_1, y_1) = (3, 2)\).

### Step 5: Calculate the distance
Substituting the values into the distance formula:
\[ D = \frac{|4(3) + 3(2) - 8|}{\sqrt{4^2 + 3^2}} \]
Calculating the numerator:
\[ = |12 + 6 - 8| = |10| = 10 \]

Calculating the denominator:
\[ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Thus, the distance \(D\) is:
\[ D = \frac{10}{5} = 2 \]

### Step 6: Determine the relationship between the line and the circle
Since the distance from the center of the circle to the line is equal to the radius of the circle (which is 2), the line is a tangent to the circle.

### Final Answer
The line \(4x + 3y - 8 = 0\) is a tangent to the circle.

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