The number of common tangents to `x^(2)+y^(2)=8, x^(2)+y^(2)=2` is

Options

A. a) one
B. b) two
C. c) four
D. d) zero

Answer

To find the number of common tangents to the circles given by the equations \(x^2 + y^2 = 8\) and \(x^2 + y^2 = 2\), we can follow these steps:

### Step 1: Identify the centers and radii of the circles
The equations of the circles can be rewritten in standard form:
1. For the first circle \(x^2 + y^2 = 8\):
- Center \(C_1 = (0, 0)\)
- Radius \(r_1 = \sqrt{8} = 2\sqrt{2}\)

2. For the second circle \(x^2 + y^2 = 2\):
- Center \(C_2 = (0, 0)\)
- Radius \(r_2 = \sqrt{2}\)

### Step 2: Determine the distance between the centers
Since both circles have their centers at the origin, the distance \(d\) between the centers \(C_1\) and \(C_2\) is:
\[ d = \sqrt{(0 - 0)^2 + (0 - 0)^2} = 0 \]

### Step 3: Analyze the relationship between the circles
Since the distance \(d = 0\) and the radius of the first circle \(r_1 = 2\sqrt{2}\) is greater than the radius of the second circle \(r_2 = \sqrt{2}\), we can conclude that the larger circle completely encompasses the smaller circle.

### Step 4: Determine the number of common tangents
When two circles have the same center and one circle is completely inside the other, there are no common tangents. Therefore, the number of common tangents is:
\[ \text{Number of common tangents} = 0 \]

### Final Answer
The number of common tangents to the circles \(x^2 + y^2 = 8\) and \(x^2 + y^2 = 2\) is **0**.

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