The number of common tangents to the circles `x^2+y^2-x = 0 and x^2 + y^2 + x = 0` are

Options

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

Answer

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

### Step 1: Rewrite the equations of the circles in standard form
We start with the equations of the circles:

1. \(x^2 + y^2 - x = 0\)
2. \(x^2 + y^2 + x = 0\)

We can rearrange these equations to make them easier to analyze.

**For the first circle:**

\[ x^2 - x + y^2 = 0 \]

To complete the square for \(x\):

\[ x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + y^2 = 0 \]

This simplifies to:

\[ \left(x - \frac{1}{2}\right)^2 + y^2 = \frac{1}{4} \]

Thus, the first circle has center \(C_1\left(\frac{1}{2}, 0\right)\) and radius \(r_1 = \frac{1}{2}\).

**For the second circle:**

\[ x^2 + x + y^2 = 0 \]

Completing the square for \(x\):

\[ x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + y^2 = 0 \]

This simplifies to:

\[ \left(x + \frac{1}{2}\right)^2 + y^2 = \frac{1}{4} \]

Thus, the second circle has center \(C_2\left(-\frac{1}{2}, 0\right)\) and radius \(r_2 = \frac{1}{2}\).

### Step 2: Calculate the distance between the centers of the circles
Now, we calculate the distance \(d\) between the centers \(C_1\) and \(C_2\):

\[ d = \sqrt{\left(\frac{1}{2} - \left(-\frac{1}{2}\right)\right)^2 + (0 - 0)^2} = \sqrt{\left(\frac{1}{2} + \frac{1}{2}\right)^2} = \sqrt{1^2} = 1 \]

### Step 3: Analyze the relationship between the distance and the radii
We have:

- \(d = 1\)
- \(r_1 + r_2 = \frac{1}{2} + \frac{1}{2} = 1\)

### Step 4: Determine the number of common tangents
According to the properties of circles:

- If \(d = r_1 + r_2\), then there are **3 common tangents** (2 external and 1 internal).

### Conclusion
Thus, the number of common tangents to the circles \(x^2 + y^2 - x = 0\) and \(x^2 + y^2 + x = 0\) is **3**.

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