The circle passing through (1, - 2) and touching the axis of x at (3,0) also passes through the point

Options

A. (-5, 2)
B. (2, -5)
C. (5, -2)
D. (-2, 5)

Answer

To solve the problem step by step, we need to find the equation of the circle that passes through the points (1, -2) and touches the x-axis at (3, 0). We will then check which of the given points lies on this circle.

### Step 1: Identify the center of the circle
Since the circle touches the x-axis at (3, 0), the center of the circle must be directly above this point. Let the center be (3, k), where k is the radius of the circle. Therefore, the radius of the circle is equal to k.

### Step 2: Write the equation of the circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our circle:
\[ (x - 3)^2 + (y - k)^2 = k^2 \]

### Step 3: Substitute the point (1, -2) into the circle's equation
Since the circle passes through the point (1, -2), we can substitute these coordinates into the equation:
\[ (1 - 3)^2 + (-2 - k)^2 = k^2 \]
Calculating this gives:
\[ (-2)^2 + (-2 - k)^2 = k^2 \]
\[ 4 + (k + 2)^2 = k^2 \]
Expanding the left side:
\[ 4 + (k^2 + 4k + 4) = k^2 \]
\[ k^2 + 4k + 8 = k^2 \]
Subtracting \(k^2\) from both sides:
\[ 4k + 8 = 0 \]
Solving for k:
\[ 4k = -8 \implies k = -2 \]

### Step 4: Substitute k back into the equation of the circle
Now that we have k, we can substitute it back into the equation of the circle:
\[ (x - 3)^2 + (y + 2)^2 = (-2)^2 \]
This simplifies to:
\[ (x - 3)^2 + (y + 2)^2 = 4 \]

### Step 5: Check which points lie on the circle
Now we need to check which of the given points satisfies the equation of the circle. We will check each point one by one.

1. **Point (-5, 2)**:
\[ (-5 - 3)^2 + (2 + 2)^2 = 64 + 16 = 80 \quad \text{(not on the circle)} \]

2. **Point (2, -5)**:
\[ (2 - 3)^2 + (-5 + 2)^2 = 1 + 9 = 10 \quad \text{(not on the circle)} \]

3. **Point (5, -2)**:
\[ (5 - 3)^2 + (-2 + 2)^2 = 4 + 0 = 4 \quad \text{(on the circle)} \]

### Conclusion
The point that lies on the circle is (5, -2).