Find the equation of a circle with centre (1,-3) and touches the line 2x-y-4=0.

Answer

To find the equation of a circle with center (1, -3) that touches the line given by the equation \(2x - y - 4 = 0\), we will follow these steps:

### Step 1: Identify the center of the circle
The center of the circle is given as \( (h, k) = (1, -3) \).

### Step 2: Find the distance from the center to the line
To find the radius of the circle, we need to calculate the distance from the center of the circle to the line. The formula to find the distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:

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

For the line \(2x - y - 4 = 0\), we can identify \(A = 2\), \(B = -1\), and \(C = -4\). The coordinates of the center are \(x_0 = 1\) and \(y_0 = -3\).

### Step 3: Substitute the values into the distance formula
Substituting the values into the distance formula:

\[ d = \frac{|2(1) + (-1)(-3) - 4|}{\sqrt{2^2 + (-1)^2}} \]

Calculating the numerator:

\[ = |2 + 3 - 4| = |1| = 1 \]

Calculating the denominator:

\[ = \sqrt{4 + 1} = \sqrt{5} \]

Thus, the distance \(d\) is:

\[ d = \frac{1}{\sqrt{5}} \]

### Step 4: Determine the radius of the circle
Since the circle touches the line, the radius \(r\) of the circle is equal to the distance from the center to the line:

\[ r = \frac{1}{\sqrt{5}} \]

### Step 5: Write the equation of the circle
The standard equation of a circle with center \((h, k)\) and radius \(r\) is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \(h = 1\), \(k = -3\), and \(r = \frac{1}{\sqrt{5}}\):

\[ (x - 1)^2 + (y + 3)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \]

Calculating \(r^2\):

\[ \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \]

Thus, the equation of the circle is:

\[ (x - 1)^2 + (y + 3)^2 = \frac{1}{5} \]

### Final Answer:
The equation of the circle is:

\[ (x - 1)^2 + (y + 3)^2 = \frac{1}{5} \]