The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle `x^(2)+y^(2)=9`, is

Options

A. (3/2, 1/2)
B. (1/2, 3/2)
C. (1/2, 1/2)
D. (1/2, `-sqrt(2)`)

Answer

To find the center of a circle that passes through the points (0, 0) and (1, 0) and touches the circle given by the equation \(x^2 + y^2 = 9\), we can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

### Step 2: Use the points (0, 0) and (1, 0)
Since the circle passes through the points (0, 0) and (1, 0), we can substitute these points into the circle's equation.

1. For the point (0, 0):
\[ (0 - h)^2 + (0 - k)^2 = r^2 \implies h^2 + k^2 = r^2 \quad \text{(1)} \]

2. For the point (1, 0):
\[ (1 - h)^2 + (0 - k)^2 = r^2 \implies (1 - h)^2 + k^2 = r^2 \]
Expanding this, we get:
\[ 1 - 2h + h^2 + k^2 = r^2 \quad \text{(2)} \]

### Step 3: Set the equations equal
From equation (1), we know that \(r^2 = h^2 + k^2\). Substituting this into equation (2):
\[ 1 - 2h + h^2 + k^2 = h^2 + k^2 \]
This simplifies to:
\[ 1 - 2h = 0 \implies h = \frac{1}{2} \quad \text{(3)} \]

### Step 4: Find the distance to the other circle
The circle \(x^2 + y^2 = 9\) has its center at (0, 0) and radius \(3\). The distance from the center of our circle \((h, k)\) to the center of the circle \(x^2 + y^2 = 9\) is given by:
\[ \sqrt{(h - 0)^2 + (k - 0)^2} = \sqrt{h^2 + k^2} \]
Since the circles are touching externally, we have:
\[ \sqrt{h^2 + k^2} = r + 3 \quad \text{(4)} \]

### Step 5: Substitute \(h\) into the equations
From equation (1) with \(h = \frac{1}{2}\):
\[ \left(\frac{1}{2}\right)^2 + k^2 = r^2 \implies \frac{1}{4} + k^2 = r^2 \quad \text{(5)} \]

### Step 6: Substitute \(r\) from equation (5) into equation (4)
From equation (5), we can express \(r\) as:
\[ r = \sqrt{\frac{1}{4} + k^2} \]
Substituting this into equation (4):
\[ \sqrt{h^2 + k^2} = \sqrt{\frac{1}{4} + k^2} + 3 \]
Squaring both sides:
\[ h^2 + k^2 = \left(\sqrt{\frac{1}{4} + k^2} + 3\right)^2 \]
Expanding gives:
\[ \frac{1}{4} + k^2 = \frac{1}{4} + k^2 + 6\sqrt{\frac{1}{4} + k^2} + 9 \]
This simplifies to:
\[ 0 = 6\sqrt{\frac{1}{4} + k^2} + 9 \]
Solving for \(k\):
\[ 0 = 6\sqrt{\frac{1}{4} + k^2} + 9 \implies k^2 = 2 \implies k = \pm \sqrt{2} \]

### Step 7: Final coordinates of the center
Thus, the center of the circle is:
\[ \left(\frac{1}{2}, \pm \sqrt{2}\right) \]

### Conclusion
The center of the circle that passes through the points (0, 0) and (1, 0) and touches the circle \(x^2 + y^2 = 9\) is:
\[ \left(\frac{1}{2}, \sqrt{2}\right) \quad \text{or} \quad \left(\frac{1}{2}, -\sqrt{2}\right) \]