The number of common tangentss to the circles `x^(2)+y^(2)-8x+2y=0 and x^(2)+y^(2)-2x-16y+25=0` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the number of common tangents to the circles given by the equations \(x^2 + y^2 - 8x + 2y = 0\) and \(x^2 + y^2 - 2x - 16y + 25 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form

1. **Circle 1 (S1)**:
\[ x^2 + y^2 - 8x + 2y = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 8x) + (y^2 + 2y) = 0 \]
\[ (x - 4)^2 - 16 + (y + 1)^2 - 1 = 0 \]
\[ (x - 4)^2 + (y + 1)^2 = 17 \]
This gives us the center \(C_1(4, -1)\) and radius \(R_1 = \sqrt{17}\).

2. **Circle 2 (S2)**:
\[ x^2 + y^2 - 2x - 16y + 25 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 2x) + (y^2 - 16y) + 25 = 0 \]
\[ (x - 1)^2 - 1 + (y - 8)^2 - 64 + 25 = 0 \]
\[ (x - 1)^2 + (y - 8)^2 = 40 \]
This gives us the center \(C_2(1, 8)\) and radius \(R_2 = \sqrt{40}\).

### Step 2: Find the distance between the centers of the circles

The distance \(d\) between the centers \(C_1(4, -1)\) and \(C_2(1, 8)\) is calculated as follows:
\[ d = \sqrt{(4 - 1)^2 + (-1 - 8)^2} = \sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} \]

### Step 3: Compare the distance with the sum of the radii

Now we calculate the sum of the radii:
\[ R_1 + R_2 = \sqrt{17} + \sqrt{40} \]
Approximating the values:
- \(\sqrt{17} \approx 4.12\)
- \(\sqrt{40} \approx 6.32\)

Thus,
\[ R_1 + R_2 \approx 4.12 + 6.32 \approx 10.44 \]

### Step 4: Analyze the relationship between \(d\) and \(R_1 + R_2\)

Now we compare:
- \(d = \sqrt{90} \approx 9.49\)
- \(R_1 + R_2 \approx 10.44\)

Since \(d < R_1 + R_2\), this indicates that the circles intersect.

### Conclusion

Since the distance between the centers \(d\) is less than the sum of the radii \(R_1 + R_2\), there are no common tangents to the circles.

### Final Answer

The number of common tangents to the circles is **0**.

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