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 with center (2, -1) that touches the line x - y - 6 = 0, we can follow these steps:

### Step 1: Identify the center of the circle
The center of the circle is given as (2, -1).

### Step 2: Determine the distance from the center to the line
To find the radius of the circle, we need to calculate the distance from the center (2, -1) to the line x - y - 6 = 0. The formula for 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 x - y - 6 = 0, we can identify \(A = 1\), \(B = -1\), and \(C = -6\).

### Step 3: Substitute the center coordinates into the distance formula
Substituting \(x_0 = 2\) and \(y_0 = -1\) into the distance formula:

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

Calculating the numerator:

\[ = |2 + 1 - 6| = |-3| = 3 \]

Calculating the denominator:

\[ = \sqrt{1 + 1} = \sqrt{2} \]

Thus, the distance \(d\) is:

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

### 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:

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

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

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

Calculating \(r^2\):

\[ \left(\frac{3\sqrt{2}}{2}\right)^2 = \frac{9 \cdot 2}{4} = \frac{18}{4} = \frac{9}{2} \]

Thus, the equation of the circle becomes:

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

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

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

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