Radius of a circle is 5. It cuts x-axis at two points at a distance 3 from the origin. Its centre is

Options

A. `(0, pm4)`
B. `(0, pm3)`
C. `(0, pm5)`
D. none

Answer

To find the center of the circle given that its radius is 5 and it intersects the x-axis at two points that are 3 units away from the origin, we can follow these steps:

### Step 1: Identify the points of intersection
The circle intersects the x-axis at points that are 3 units away from the origin. Therefore, the points of intersection are:
- (-3, 0)
- (3, 0)

### Step 2: Write the general equation of the circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Given that the radius \( r = 5 \), we can write the equation as:
\[ (x - h)^2 + (y - k)^2 = 25 \]

### Step 3: Substitute the points into the equation
Since the circle passes through the points (-3, 0) and (3, 0), we can substitute these points into the circle's equation to find h and k.

#### For point (-3, 0):
Substituting (-3, 0) into the equation:
\[ (-3 - h)^2 + (0 - k)^2 = 25 \]
This simplifies to:
\[ (-3 - h)^2 + k^2 = 25 \tag{1} \]

#### For point (3, 0):
Substituting (3, 0) into the equation:
\[ (3 - h)^2 + (0 - k)^2 = 25 \]
This simplifies to:
\[ (3 - h)^2 + k^2 = 25 \tag{2} \]

### Step 4: Set up the equations
Now we have two equations:
1. \((-3 - h)^2 + k^2 = 25\)
2. \((3 - h)^2 + k^2 = 25\)

### Step 5: Expand both equations
Expanding equation (1):
\[ (-3 - h)^2 = 9 + 6h + h^2 \]
So, equation (1) becomes:
\[ 9 + 6h + h^2 + k^2 = 25 \]
This simplifies to:
\[ 6h + h^2 + k^2 = 16 \tag{3} \]

Expanding equation (2):
\[ (3 - h)^2 = 9 - 6h + h^2 \]
So, equation (2) becomes:
\[ 9 - 6h + h^2 + k^2 = 25 \]
This simplifies to:
\[ -6h + h^2 + k^2 = 16 \tag{4} \]

### Step 6: Solve the equations
Now we have two equations (3) and (4):
1. \(6h + h^2 + k^2 = 16\)
2. \(-6h + h^2 + k^2 = 16\)

Subtract equation (4) from equation (3):
\[ (6h + h^2 + k^2) - (-6h + h^2 + k^2) = 0 \]
This simplifies to:
\[ 12h = 0 \implies h = 0 \]

### Step 7: Substitute h back to find k
Substituting \( h = 0 \) into equation (3):
\[ 6(0) + (0)^2 + k^2 = 16 \]
This simplifies to:
\[ k^2 = 16 \implies k = \pm 4 \]

### Step 8: Conclusion
The center of the circle is at the points:
\[ (0, 4) \text{ or } (0, -4) \]

### Final Answer
The center of the circle is \( (0, 4) \) or \( (0, -4) \).

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