Let the equation of circle is `x ^(2) + y ^(2) - 6x + 4y + 12=0.` Then the line `3x - 4y + 5` is a

Options

A. chord of given cicel
B. tangent of given circle
C. diameter of given circle
D. does not met the given circle at all

Answer

To determine the relationship between the given circle and the line, we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 4y + 12 = 0 \]

We can rearrange this equation to identify the center and radius. First, we group the \(x\) and \(y\) terms:
\[ (x^2 - 6x) + (y^2 + 4y) + 12 = 0 \]

Next, we complete the square for the \(x\) and \(y\) terms.

**Completing the square for \(x\):**
1. Take half of the coefficient of \(x\) (which is -6), square it: \((-6/2)^2 = 9\).
2. Add and subtract 9 inside the equation.

**Completing the square for \(y\):**
1. Take half of the coefficient of \(y\) (which is 4), square it: \((4/2)^2 = 4\).
2. Add and subtract 4 inside the equation.

Now, we rewrite the equation:
\[ (x^2 - 6x + 9) + (y^2 + 4y + 4) - 9 - 4 + 12 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y + 2)^2 - 1 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x - 3)^2 + (y + 2)^2 = 1 \]

### Step 2: Identify the center and radius of the circle
From the standard form of the circle:
- Center \(C = (3, -2)\)
- Radius \(R = \sqrt{1} = 1\)

### Step 3: Find the distance from the center of the circle to the line
The equation of the line is given as:
\[ 3x - 4y + 5 = 0 \]

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

Here, \(A = 3\), \(B = -4\), \(C = 5\), and the center of the circle \((x_1, y_1) = (3, -2)\).

Substituting these values into the distance formula:
\[ d = \frac{|3(3) - 4(-2) + 5|}{\sqrt{3^2 + (-4)^2}} \]
Calculating the numerator:
\[ = |9 + 8 + 5| = |22| = 22 \]
Calculating the denominator:
\[ = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus, the distance \(d\) is:
\[ d = \frac{22}{5} = 4.4 \]

### Step 4: Compare the distance with the radius
We found that the radius \(R = 1\) and the distance \(d = 4.4\).

Since \(d > R\) (4.4 > 1), this means that the line does not intersect the circle at all.

### Conclusion
The line \(3x - 4y + 5 = 0\) is an external line to the circle.

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