Show that 3x - 4y - 11 = 0 is a tangent to the circle `x^(2) + y^(2) - 8y + 15 = 0` and find the equation of the other tangent which is parallel to the st. line 3x = 4y.

Answer

To solve the problem, we need to show that the line \(3x - 4y - 11 = 0\) is a tangent to the circle given by the equation \(x^2 + y^2 - 8y + 15 = 0\) and then find the equation of another tangent that is parallel to the line \(3x = 4y\).

### Step 1: Rewrite the Circle's Equation
First, we rewrite the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 - 8y + 15 = 0 \]

We can complete the square for the \(y\) terms:

\[ x^2 + (y^2 - 8y) + 15 = 0 \]
\[ x^2 + (y - 4)^2 - 16 + 15 = 0 \]
\[ x^2 + (y - 4)^2 - 1 = 0 \]
\[ x^2 + (y - 4)^2 = 1 \]

From this, we can see that the center of the circle is \((0, 4)\) and the radius is \(1\).

### Step 2: Find the Distance from the Center to the Line
Next, we need to find the distance from the center of the circle \((0, 4)\) to the line \(3x - 4y - 11 = 0\). The formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our line, \(A = 3\), \(B = -4\), and \(C = -11\). The coordinates of the center are \((0, 4)\).

Substituting these values into the distance formula:

\[ d = \frac{|3(0) - 4(4) - 11|}{\sqrt{3^2 + (-4)^2}} = \frac{|0 - 16 - 11|}{\sqrt{9 + 16}} = \frac{|-27|}{5} = \frac{27}{5} \]

### Step 3: Check if the Line is a Tangent
For the line to be a tangent to the circle, the distance from the center to the line must equal the radius of the circle. The radius of the circle is \(1\).

Since \(\frac{27}{5} \neq 1\), we need to check our calculations.

### Step 4: Correct Calculation of Distance
Let's re-evaluate the distance calculation:

\[ d = \frac{|3(0) - 4(4) - 11|}{\sqrt{3^2 + (-4)^2}} = \frac{|0 - 16 - 11|}{\sqrt{9 + 16}} = \frac{|-27|}{5} = \frac{27}{5} \]

This indicates that the line is not a tangent. However, we need to check if the distance equals the radius.

### Step 5: Finding the Other Tangent
Since the line \(3x = 4y\) can be rewritten as \(y = \frac{3}{4}x\), we know that the slope of this line is \(\frac{3}{4}\). The other tangent line will also have this slope.

The equation of a line with slope \(\frac{3}{4}\) can be expressed as:

\[ y - y_1 = m(x - x_1) \]

Using the point \((0, 4)\) (the center of the circle) and the slope \(\frac{3}{4}\):

\[ y - 4 = \frac{3}{4}(x - 0) \]
\[ y = \frac{3}{4}x + 4 \]

### Step 6: Finding the Parallel Tangent
To find the equation of the other tangent that is parallel to the line \(3x = 4y\), we can write it in the form \(y = \frac{3}{4}x + c\).

To find \(c\), we need the distance from the center to this new line to equal the radius \(1\). The distance from the center \((0, 4)\) to the line \(y = \frac{3}{4}x + c\) is given by:

\[ d = \frac{|-\frac{3}{4}(0) + 1(4) - c|}{\sqrt{(\frac{3}{4})^2 + 1^2}} = \frac{|4 - c|}{\sqrt{\frac{9}{16} + 1}} = \frac{|4 - c|}{\sqrt{\frac{25}{16}}} = \frac{|4 - c|}{\frac{5}{4}} = \frac{4 - c}{\frac{5}{4}} = \frac{4 - c}{5} \cdot 4 \]

Setting this equal to \(1\):

\[ \frac{|4 - c|}{5} = 1 \]
\[ |4 - c| = 5 \]

This gives us two equations:

1. \(4 - c = 5 \Rightarrow c = -1\)
2. \(4 - c = -5 \Rightarrow c = 9\)

Thus, the equations of the tangents are:

1. \(y = \frac{3}{4}x - 1\)
2. \(y = \frac{3}{4}x + 9\)

### Final Answer
The other tangent line is \(y = \frac{3}{4}x - 1\) or \(y = \frac{3}{4}x + 9\).