find the equations of tangents to the circle `x^2+y^2-4x-6y-12=0` which are perpendicular to the line `4x+3y=7`

Answer

To find the equations of the tangents to the circle given by the equation \(x^2 + y^2 - 4x - 6y - 12 = 0\) that are perpendicular to the line \(4x + 3y = 7\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we will rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]
We can rearrange it:
\[ x^2 - 4x + y^2 - 6y = 12 \]

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).

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

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

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 12 \]
\[ (x - 2)^2 + (y - 3)^2 = 25 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center: \((h, k) = (2, 3)\)
- Radius: \(r = \sqrt{25} = 5\)

### Step 4: Determine the Slope of the Given Line
The line \(4x + 3y = 7\) can be rewritten 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 5: Find the Slope of the Tangents
Since the tangents are perpendicular to the line, the slope of the tangents will be the negative reciprocal of \(-\frac{4}{3}\):
\[ \text{slope of tangent} = \frac{3}{4} \]

### Step 6: Write the Equation of the Tangent Line
The equation of a line with slope \(m\) passing through a point \((x_0, y_0)\) is given by:
\[ y - y_0 = m(x - x_0) \]
Using the center \((2, 3)\) and the slope \(\frac{3}{4}\):
\[ y - 3 = \frac{3}{4}(x - 2) \]
This simplifies to:
\[ y - 3 = \frac{3}{4}x - \frac{3}{2} \]
\[ y = \frac{3}{4}x + \frac{3}{2} \]

### Step 7: General Form of the Tangent Line
To express this in the standard form \(Ax + By + C = 0\):
\[ 3x - 4y + 6 = 0 \]

### Step 8: Use the Distance Formula
We need to find the values of \(c\) such that the distance from the center of the circle to the tangent line equals the radius (5). The general form of the tangent line is:
\[ 3x - 4y + c = 0 \]
Using the distance formula from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \(A = 3\), \(B = -4\), \(C = c\), and \((x_0, y_0) = (2, 3)\):
\[ d = \frac{|3(2) - 4(3) + c|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 12 + c|}{5} = \frac{|c - 6|}{5} \]
Setting this equal to the radius:
\[ \frac{|c - 6|}{5} = 5 \]
\[ |c - 6| = 25 \]

### Step 9: Solve for \(c\)
This gives us two cases:
1. \(c - 6 = 25 \implies c = 31\)
2. \(c - 6 = -25 \implies c = -19\)

### Step 10: Write the Tangent Equations
Substituting \(c\) back into the tangent line equation:
1. For \(c = 31\):
\[ 3x - 4y + 31 = 0 \]
2. For \(c = -19\):
\[ 3x - 4y - 19 = 0 \]

### Final Answer
The equations of the tangents to the circle that are perpendicular to the line \(4x + 3y = 7\) are:
1. \(3x - 4y + 31 = 0\)
2. \(3x - 4y - 19 = 0\)