Show that the line 3x-4y=1 touches the circle `x^(2)+y^(2)-2x+4y+1=0`. Find the coordinates of the point of contact.

Answer

To show that the line \(3x - 4y = 1\) touches the circle given by the equation \(x^2 + y^2 - 2x + 4y + 1 = 0\), we will follow these steps:

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

\[ x^2 + y^2 - 2x + 4y + 1 = 0 \]

We can complete the square for \(x\) and \(y\).

For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

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

Substituting these into the circle equation gives:

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

Simplifying this, we have:

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

Thus, we can write:

\[ (x - 1)^2 + (y + 2)^2 = 4 \]

This represents a circle with center \((1, -2)\) and radius \(2\).

### Step 2: Rewrite the Line Equation
Next, we rewrite the line equation \(3x - 4y = 1\) in slope-intercept form:

\[ 4y = 3x - 1 \quad \Rightarrow \quad y = \frac{3}{4}x - \frac{1}{4} \]

### Step 3: Find the Distance from the Center of the Circle to the Line
To determine if the line is tangent to the circle, we calculate the distance \(d\) from the center of the circle \((1, -2)\) to the line \(3x - 4y - 1 = 0\) using the formula:

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

where \(A = 3\), \(B = -4\), \(C = -1\), and \((x_0, y_0) = (1, -2)\).

Calculating \(d\):

\[ d = \frac{|3(1) - 4(-2) - 1|}{\sqrt{3^2 + (-4)^2}} = \frac{|3 + 8 - 1|}{\sqrt{9 + 16}} = \frac{|10|}{\sqrt{25}} = \frac{10}{5} = 2 \]

### Step 4: Compare the Distance with the Radius
The radius of the circle is \(2\). Since the distance \(d = 2\) is equal to the radius, the line is tangent to the circle.

### Step 5: Find the Point of Contact
To find the coordinates of the point of contact, we can substitute \(y = \frac{3}{4}x - \frac{1}{4}\) into the circle's equation:

Substituting \(y\) into the circle equation:

\[ (x - 1)^2 + \left(\frac{3}{4}x - \frac{1}{4} + 2\right)^2 = 4 \]

Simplifying the second term:

\[ \frac{3}{4}x - \frac{1}{4} + 2 = \frac{3}{4}x - \frac{1}{4} + \frac{8}{4} = \frac{3}{4}x + \frac{7}{4} \]

Thus, we have:

\[ (x - 1)^2 + \left(\frac{3}{4}x + \frac{7}{4}\right)^2 = 4 \]

Expanding this gives:

\[ (x - 1)^2 + \left(\frac{3x + 7}{4}\right)^2 = 4 \]

Now, squaring the second term:

\[ (x - 1)^2 + \frac{(3x + 7)^2}{16} = 4 \]

Now, we can solve for \(x\) and subsequently find \(y\).

After solving, we find:

\[ x = -\frac{1}{5} \]

Substituting back to find \(y\):

\[ y = \frac{3(-\frac{1}{5}) - 1}{4} = -\frac{3}{20} - \frac{5}{20} = -\frac{8}{20} = -\frac{2}{5} \]

Thus, the point of contact is:

\[ \left(-\frac{1}{5}, -\frac{2}{5}\right) \]

### Final Answer
The line \(3x - 4y = 1\) touches the circle \(x^2 + y^2 - 2x + 4y + 1 = 0\) at the point \(\left(-\frac{1}{5}, -\frac{2}{5}\right)\).