Let the point at which the circle passing through (0, 0) and (1, 0) touches the circle `x^(2)+y^(2)=9` is P(h, k), then `|k|` is equal to

Options

A. `sqrt5`
B. `2sqrt2`
C. `sqrt6`
D. `sqrt7`

Answer

To solve the problem step by step, we need to find the value of \(|k|\) where \(P(h, k)\) is the point at which the circle passing through \((0, 0)\) and \((1, 0)\) touches the circle defined by the equation \(x^2 + y^2 = 9\).

### Step 1: Understand the circles involved
The first circle is defined by the equation \(x^2 + y^2 = 9\). This circle has:
- Center at \((0, 0)\)
- Radius \(r_2 = 3\)

The second circle passes through the points \((0, 0)\) and \((1, 0)\). Its general equation can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 2: Find the parameters of the second circle
Since the second circle passes through the origin \((0, 0)\), substituting these coordinates into the equation gives:
\[ 0 + 0 + 0 + 0 + c = 0 \implies c = 0 \]
Thus, the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0 \]

Next, substituting the point \((1, 0)\):
\[ 1^2 + 0 + 2g(1) + 0 = 0 \implies 1 + 2g = 0 \implies g = -\frac{1}{2} \]
So, the equation of the second circle now becomes:
\[ x^2 + y^2 - x + 2fy = 0 \]

### Step 3: Identify the center of the second circle
The center of the second circle is given by \((-g, -f)\):
\[ \text{Center} = \left(\frac{1}{2}, -f\right) \]

### Step 4: Use the condition of tangency
The two circles touch each other at point \(P(h, k)\). The distance between the centers of the two circles must equal the difference of their radii:
\[ \text{Distance} = r_2 - r_1 \]
Where \(r_1\) is the radius of the second circle. The radius \(r_1\) can be expressed as:
\[ r_1 = \sqrt{g^2 + f^2} \]
Substituting \(g = -\frac{1}{2}\):
\[ r_1 = \sqrt{\left(-\frac{1}{2}\right)^2 + f^2} = \sqrt{\frac{1}{4} + f^2} \]

The distance between the centers is:
\[ \sqrt{\left(\frac{1}{2} - 0\right)^2 + (-f - 0)^2} = \sqrt{\frac{1}{4} + f^2} \]

Setting the distances equal gives:
\[ \sqrt{\frac{1}{4} + f^2} = 3 - r_1 \]

### Step 5: Solve for \(f\)
Squaring both sides:
\[ \frac{1}{4} + f^2 = (3 - r_1)^2 \]
Substituting \(r_1\):
\[ \frac{1}{4} + f^2 = \left(3 - \sqrt{\frac{1}{4} + f^2}\right)^2 \]

### Step 6: Solve the equations
This leads to a quadratic equation in \(f\). After solving, we find \(f = -\sqrt{2}\) (since \(f\) must be negative for the center to be below the x-axis).

### Step 7: Find \(|k|\)
The coordinates of point \(P(h, k)\) must satisfy both circle equations. Substituting \(f\) back into the equations will yield \(k\). After solving, we find:
\[ k = 2\sqrt{2} \]
Thus, \(|k| = 2\sqrt{2}\).

### Final Answer
\[ |k| = 2\sqrt{2} \]