The number of common tangents to the circles one of which passes through the origin and cuts off intercepts 2 from each of the axes and the other circle has the segment joining the origin and the point (1, 1) as a diameter, is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the number of common tangents to the two given circles, we will follow these steps:

### Step 1: Determine the equation of Circle 1
Circle 1 passes through the origin (0,0) and cuts off intercepts of 2 from each axis. The intercepts are at points (2,0) and (0,2).

Using the general equation of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Substituting the point (0,0):
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0 \]

Now substituting the point (2,0):
\[ 2^2 + 0^2 + 2g(2) + 2f(0) + 0 = 0 \implies 4 + 4g = 0 \implies g = -1 \]

Next, substituting the point (0,2):
\[ 0^2 + 2^2 + 2g(0) + 2f(2) + 0 = 0 \implies 4 + 4f = 0 \implies f = -1 \]

Thus, the equation of Circle 1 is:
\[ x^2 + y^2 - 2x - 2y = 0 \]

### Step 2: Determine the center and radius of Circle 1
The center \(C_1\) of Circle 1 is given by \((-g, -f)\):
\[ C_1 = (1, 1) \]
The radius \(r_1\) is given by:
\[ r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-1)^2 - 0} = \sqrt{2} \]

### Step 3: Determine the equation of Circle 2
Circle 2 has the segment joining the origin (0,0) and the point (1,1) as its diameter. The center \(C_2\) is the midpoint of the segment:
\[ C_2 = \left(\frac{0+1}{2}, \frac{0+1}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right) \]
The radius \(r_2\) is half the distance between (0,0) and (1,1):
\[ r_2 = \frac{1}{2} \sqrt{(1-0)^2 + (1-0)^2} = \frac{1}{2} \sqrt{2} = \frac{\sqrt{2}}{2} \]

The equation of Circle 2 is:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{\sqrt{2}}{2}\right)^2 \]
Expanding this gives:
\[ x^2 + y^2 - x - y = 0 \]

### Step 4: Find the distance between the centers
The distance \(d\) between the centers \(C_1\) and \(C_2\) is calculated as follows:
\[ d = \sqrt{\left(1 - \frac{1}{2}\right)^2 + \left(1 - \frac{1}{2}\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Compare the distance with the radii
Now, we compare the distance \(d\) with the difference of the radii:
\[ r_1 - r_2 = \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \]

### Conclusion
Since the distance \(d\) is equal to the difference of the radii \(r_1 - r_2\), the circles touch each other internally. Therefore, there is exactly **1 common tangent**.

### Final Answer
The number of common tangents to the circles is **1**.

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