Find the equation of circle with Centre C (1,- 3) and tangent to 2 x -y - 4 = 0.

Answer

To find the equation of the circle with center \( C(1, -3) \) and tangent to the line \( 2x - y - 4 = 0 \), we can follow these steps:

### Step 1: Identify the center and the equation of the tangent line
The center of the circle is given as \( C(1, -3) \). The equation of the tangent line is \( 2x - y - 4 = 0 \).

### Step 2: Calculate the perpendicular distance from the center to the tangent line
The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 2 \), \( B = -1 \), and \( C = -4 \). The coordinates of the center are \( (x_0, y_0) = (1, -3) \).

Substituting these values into the formula:
\[ d = \frac{|2(1) + (-1)(-3) - 4|}{\sqrt{2^2 + (-1)^2}} \]
\[ = \frac{|2 + 3 - 4|}{\sqrt{4 + 1}} = \frac{|1|}{\sqrt{5}} = \frac{1}{\sqrt{5}} \]

### Step 3: Determine the radius of the circle
Since the circle is tangent to the line, the radius \( r \) of the circle is equal to the perpendicular distance \( d \):
\[ r = \frac{1}{\sqrt{5}} \]

### Step 4: Write the equation of the circle
The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 1 \), \( k = -3 \), and \( r = \frac{1}{\sqrt{5}} \):
\[ (x - 1)^2 + (y + 3)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \]
\[ (x - 1)^2 + (y + 3)^2 = \frac{1}{5} \]

### Final Equation
Thus, the equation of the circle is:
\[ (x - 1)^2 + (y + 3)^2 = \frac{1}{5} \]

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