Equation of circle touching the lines `|x-2| + | y-3|=4` will be

Options

A. `(x-2)^(2)+ (y-3)^(2) =12`
B. `(x-2)^(2) + (y-3)^(2) =4`
C. `(x-2)^(2) + (y -3)^(2) =4`
D. `(x-2)^(2) + (y-3)^(2) =8`

Answer

To find the equation of the circle that touches the lines given by the equation \(|x - 2| + |y - 3| = 4\), we will follow these steps:

### Step 1: Understand the given equation
The equation \(|x - 2| + |y - 3| = 4\) represents a square in the coordinate plane. The center of this square is at the point \((2, 3)\).

### Step 2: Determine the vertices of the square
The vertices of the square can be found by considering the four cases of the absolute value equation:
1. \(x - 2 + y - 3 = 4\) → \(x + y = 9\)
2. \(x - 2 - (y - 3) = 4\) → \(x - y = 3\)
3. \(-(x - 2) + (y - 3) = 4\) → \(-x + y = 5\) → \(x - y = -5\)
4. \(-(x - 2) - (y - 3) = 4\) → \(-x - y = 1\) → \(x + y = -1\)

The vertices of the square can be calculated by solving these equations pairwise.

### Step 3: Calculate the radius of the circle
The radius of the circle will be the distance from the center of the square \((2, 3)\) to any of the sides of the square. We can take the distance to one of the lines, for example, the line \(x + y = 9\).

Using the formula for the distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\):
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(x + y - 9 = 0\) (where \(A = 1\), \(B = 1\), and \(C = -9\)):
\[ \text{Distance} = \frac{|1 \cdot 2 + 1 \cdot 3 - 9|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 9|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 4: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, the center \((h, k) = (2, 3)\) and the radius \(r = 2\sqrt{2}\). Therefore:
\[ (x - 2)^2 + (y - 3)^2 = (2\sqrt{2})^2 = 8 \]

### Final Answer
The equation of the circle that touches the lines is:
\[ (x - 2)^2 + (y - 3)^2 = 8 \]

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