If the two straight lines `3x - 2y - 8=0` and `2x - y -5=0` lie along two diameters of a circle which touches the x-axis then the equation of the circle is

Options

A. `(x-2)^(2)+(y-1)^(2)=1`
B. `(x+2)^(2) +(y-1)^(2)=1`
C. `(x-2)^(2) +(y+1)^(2) =1`
D. none

Answer

To find the equation of the circle that touches the x-axis and has its diameters along the given lines, we can follow these steps:

### Step 1: Find the intersection point of the two lines
We need to solve the equations of the two lines to find their intersection point, which will be the center of the circle.

Given lines:
1. \(3x - 2y - 8 = 0\) (Equation 1)
2. \(2x - y - 5 = 0\) (Equation 2)

**Solving the equations:**

From Equation 2, we can express \(y\) in terms of \(x\):
\[ y = 2x - 5 \]

Now, substitute this value of \(y\) into Equation 1:
\[ 3x - 2(2x - 5) - 8 = 0 \]
\[ 3x - 4x + 10 - 8 = 0 \]
\[ -x + 2 = 0 \implies x = 2 \]

Now, substitute \(x = 2\) back into the expression for \(y\):
\[ y = 2(2) - 5 = 4 - 5 = -1 \]

Thus, the intersection point (center of the circle) is:
\[ C(2, -1) \]

### Step 2: Determine the radius of the circle
Since the circle touches the x-axis, the radius of the circle is equal to the distance from the center to the x-axis. The y-coordinate of the center is -1, so the radius \(r\) is:
\[ r = |-1| = 1 \]

### Step 3: Write the equation of the circle
The general 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 \(r = 1\):
\[ (x - 2)^2 + (y + 1)^2 = 1^2 \]
\[ (x - 2)^2 + (y + 1)^2 = 1 \]

### Final Answer
The equation of the circle is:
\[ (x - 2)^2 + (y + 1)^2 = 1 \]

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