Find the equation of a circle whose centre is (2,-1) and touches the line x-y-6=0.

Answer

To find the equation of a circle whose center is at (2, -1) and touches the line given by the equation \(x - y - 6 = 0\), we can follow these steps:

### Step 1: Identify the center and the line
The center of the circle is given as \(C(2, -1)\). The line can be rewritten in the standard form \(Ax + By + C = 0\) where \(A = 1\), \(B = -1\), and \(C = -6\).

### Step 2: Calculate the perpendicular distance from the center to the line
The formula for the perpendicular distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting the values \(A = 1\), \(B = -1\), \(C = -6\), and the center coordinates \(x_1 = 2\) and \(y_1 = -1\):

\[ d = \frac{|1(2) + (-1)(-1) - 6|}{\sqrt{1^2 + (-1)^2}} = \frac{|2 + 1 - 6|}{\sqrt{1 + 1}} = \frac{|-3|}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]

### Step 3: Determine the radius of the circle
Since the circle touches the line, the radius \(r\) of the circle is equal to the perpendicular distance calculated in Step 2:

\[ r = \frac{3}{\sqrt{2}} \]

### Step 4: 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 \(h = 2\), \(k = -1\), and \(r = \frac{3}{\sqrt{2}}\):

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

Calculating \(r^2\):

\[ \left(\frac{3}{\sqrt{2}}\right)^2 = \frac{9}{2} \]

Thus, the equation of the circle becomes:

\[ (x - 2)^2 + (y + 1)^2 = \frac{9}{2} \]

### Step 5: Expand the equation
Expanding the equation:

\[ (x - 2)^2 + (y + 1)^2 = \frac{9}{2} \]

\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = \frac{9}{2} \]

Combining terms:

\[ x^2 + y^2 - 4x + 2y + 5 = \frac{9}{2} \]

Multiplying through by 2 to eliminate the fraction:

\[ 2x^2 + 2y^2 - 8x + 4y + 10 = 9 \]

Rearranging gives:

\[ 2x^2 + 2y^2 - 8x + 4y + 1 = 0 \]

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

\[ 2x^2 + 2y^2 - 8x + 4y + 1 = 0 \]