The piont where the line 4x - 3y + 7 = 0 touches the circle `x^(2) + y^(2) - 6x + 4y - 12 = 0` is

Options

A. a) (1,1)
B. b)(1,-1)
C. c)(-1,1)
D. d)(-1,-1)

Answer

To find the point where the line \(4x - 3y + 7 = 0\) touches the circle \(x^2 + y^2 - 6x + 4y - 12 = 0\), 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 complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]

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

Substituting these into the circle's equation gives:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \]
\[ (x - 3)^2 + (y + 2)^2 - 25 = 0 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
This shows that the center of the circle is \((3, -2)\) and the radius is \(5\).

### Step 2: Find the slope of the tangent line
The line given is \(4x - 3y + 7 = 0\). We can rewrite it in slope-intercept form:
\[ 3y = 4x + 7 \implies y = \frac{4}{3}x + \frac{7}{3} \]
The slope of this line is \(\frac{4}{3}\).

### Step 3: Find the perpendicular slope
The slope of the radius at the point of tangency will be the negative reciprocal of the slope of the tangent line:
\[ \text{slope of radius} = -\frac{3}{4} \]

### Step 4: Write the equation of the radius
Using the center of the circle \((3, -2)\) and the slope \(-\frac{3}{4}\), we can write the equation of the radius:
\[ y + 2 = -\frac{3}{4}(x - 3) \]
Multiplying through by \(4\) to eliminate the fraction:
\[ 4(y + 2) = -3(x - 3) \]
\[ 4y + 8 = -3x + 9 \]
\[ 3x + 4y - 1 = 0 \]

### Step 5: Find the intersection of the tangent line and the radius
Now we have two equations:
1. \(4x - 3y + 7 = 0\) (tangent line)
2. \(3x + 4y - 1 = 0\) (radius)

We can solve this system of equations. From the first equation, we can express \(y\):
\[ 3y = 4x + 7 \implies y = \frac{4}{3}x + \frac{7}{3} \]
Substituting this into the second equation:
\[ 3x + 4\left(\frac{4}{3}x + \frac{7}{3}\right) - 1 = 0 \]
\[ 3x + \frac{16}{3}x + \frac{28}{3} - 1 = 0 \]
Multiplying through by \(3\) to eliminate the fraction:
\[ 9x + 16x + 28 - 3 = 0 \]
\[ 25x + 25 = 0 \]
\[ x = -1 \]

### Step 6: Find the corresponding \(y\) value
Substituting \(x = -1\) back into the equation of the tangent line:
\[ 4(-1) - 3y + 7 = 0 \]
\[ -4 - 3y + 7 = 0 \]
\[ 3y = 3 \implies y = 1 \]

### Final Answer
The point where the line touches the circle is \((-1, 1)\).