Find the equations of the tangents to the circle `x^(2)+y^(2)-22x-4y+25=0` and perpendicular to the line `5x+12y+9=0`

Answer

To find the equations of the tangents to the circle given by the equation \(x^2 + y^2 - 22x - 4y + 25 = 0\) that are perpendicular to the line \(5x + 12y + 9 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 22x - 4y + 25 = 0 \]
We can complete the square for \(x\) and \(y\).

For \(x\):
\[ x^2 - 22x = (x - 11)^2 - 121 \]

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

Substituting these back into the equation gives:
\[ (x - 11)^2 - 121 + (y - 2)^2 - 4 + 25 = 0 \]
Simplifying this, we have:
\[ (x - 11)^2 + (y - 2)^2 - 100 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x - 11)^2 + (y - 2)^2 = 100 \]
This shows that the center of the circle is \((11, 2)\) and the radius is \(10\).

### Step 2: Find the Slope of the Given Line
Next, we find the slope of the line \(5x + 12y + 9 = 0\). Rearranging this into slope-intercept form \(y = mx + b\):
\[ 12y = -5x - 9 \implies y = -\frac{5}{12}x - \frac{3}{4} \]
The slope \(m_1\) of the line is \(-\frac{5}{12}\).

### Step 3: Find the Slope of the Tangent
The slope of the tangent line \(m_2\) that is perpendicular to the given line can be found using the negative reciprocal:
\[ m_2 = \frac{12}{5} \]

### Step 4: Use the Point-Slope Form of the Tangent Line
The equation of the tangent line can be expressed using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is the center of the circle \((11, 2)\) and \(m = \frac{12}{5}\):
\[ y - 2 = \frac{12}{5}(x - 11) \]
Rearranging this gives:
\[ y = \frac{12}{5}x - \frac{132}{5} + 2 \]
\[ y = \frac{12}{5}x - \frac{132}{5} + \frac{10}{5} \]
\[ y = \frac{12}{5}x - \frac{122}{5} \]

### Step 5: General Form of the Tangent Line
To convert this into general form:
\[ 12x - 5y - 122 = 0 \]

### Step 6: Find the Second Tangent
Now we will find the second tangent line. The distance from the center of the circle to the line must equal the radius. The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \(A = 5\), \(B = 12\), \(C = c\), and the center \((11, 2)\):
\[ d = \frac{|5(11) + 12(2) + c|}{\sqrt{5^2 + 12^2}} = 10 \]
Calculating the denominator:
\[ \sqrt{25 + 144} = \sqrt{169} = 13 \]
Thus:
\[ \frac{|55 + 24 + c|}{13} = 10 \]
This simplifies to:
\[ |79 + c| = 130 \]
This gives two equations:
1. \(79 + c = 130 \implies c = 51\)
2. \(79 + c = -130 \implies c = -209\)

### Step 7: Write the Tangent Equations
Using these values of \(c\):
1. For \(c = 51\):
\[ 5y - 12x - 51 = 0 \]
2. For \(c = -209\):
\[ 5y - 12x + 209 = 0 \]

### Final Answer
The equations of the tangents to the circle that are perpendicular to the given line are:
1. \(5y - 12x - 51 = 0\)
2. \(5y - 12x + 209 = 0\)