Find the number of common tangents of two circles `x^(2)+y^(2)=4,x^(2)+y^(2)-6x-8y+16=0`,

Answer

To find the number of common tangents of the two circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 6x - 8y + 16 = 0\), we will follow these steps:

### Step 1: Identify the first circle
The first circle is given by the equation:
\[ x^2 + y^2 = 4 \]
This can be rewritten in standard form as:
\[ (x - 0)^2 + (y - 0)^2 = 2^2 \]
From this, we can identify:
- Center \(C_1 = (0, 0)\)
- Radius \(r_1 = 2\)

### Step 2: Identify the second circle
The second circle is given by the equation:
\[ x^2 + y^2 - 6x - 8y + 16 = 0 \]
We can rearrange this into standard form by completing the square:
\[ (x^2 - 6x) + (y^2 - 8y) + 16 = 0 \]
Completing the square for \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Completing the square for \(y\):
\[ y^2 - 8y = (y - 4)^2 - 16 \]
Substituting these back, we have:
\[ ((x - 3)^2 - 9) + ((y - 4)^2 - 16) + 16 = 0 \]
Simplifying gives:
\[ (x - 3)^2 + (y - 4)^2 - 9 = 0 \]
Thus, we can rewrite it as:
\[ (x - 3)^2 + (y - 4)^2 = 3^2 \]
From this, we identify:
- Center \(C_2 = (3, 4)\)
- Radius \(r_2 = 3\)

### Step 3: Calculate the distance between the centers
Now, we need to calculate the distance \(d\) between the centers \(C_1\) and \(C_2\):
\[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Determine the number of common tangents
To find the number of common tangents, we compare \(d\) with \(r_1 + r_2\) and \(|r_1 - r_2|\):
- \(r_1 + r_2 = 2 + 3 = 5\)
- \(|r_1 - r_2| = |2 - 3| = 1\)

Since \(d = r_1 + r_2\), the circles are externally tangent to each other. When two circles are externally tangent, they have exactly one common tangent.

### Conclusion
The number of common tangents of the two circles is **1**.

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