Centre of a circle is (2, 3). If the line `x+y=1` touches it, its equation is

Options

A. `x^(2)+y^(2)-4x-6y+4=0`
B. `x^(2)+y^(2) -4x-6y+5=0`
C. `x^(2)+y^(2)-4x-6y-5=0`
D. none of these

Answer

To find the equation of the circle with center (2, 3) that touches the line \(x + y = 1\), we can follow these steps:

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

### Step 2: Find the distance from the center to the line
The distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by the formula:

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

For the line \(x + y = 1\), we can rewrite it in the form \(Ax + By + C = 0\):

\[ x + y - 1 = 0 \quad \Rightarrow \quad A = 1, B = 1, C = -1 \]

Now substituting the center coordinates \((x_0, y_0) = (2, 3)\):

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

### Step 3: Determine the radius of the circle
Since the line touches the circle, the distance we calculated is the radius \(r\) of the circle. Thus, \(r = 2\sqrt{2}\).

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

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

Calculating \(r^2\):

\[ (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

So the equation becomes:

\[ (x - 2)^2 + (y - 3)^2 = 8 \]

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

\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) = 8 \]

Combining like terms:

\[ x^2 + y^2 - 4x - 6y + 13 = 8 \]

Rearranging gives:

\[ x^2 + y^2 - 4x - 6y + 5 = 0 \]

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

\[ x^2 + y^2 - 4x - 6y + 5 = 0 \]

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