Find the transverse common tangents of the circles `x^(2) + y^(2) -4x -10y + 28 = 0 ` and `x^(2) + y^(2) + 4x - 6y + 4= 0. `

Answer

To find the transverse common tangents of the circles given by the equations:

1. \( x^2 + y^2 - 4x - 10y + 28 = 0 \)
2. \( x^2 + y^2 + 4x - 6y + 4 = 0 \)

we will follow these steps:

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

For the first circle, we can rearrange the equation:

\[ x^2 - 4x + y^2 - 10y + 28 = 0 \]

Completing the square for \(x\) and \(y\):

\[ (x^2 - 4x + 4) + (y^2 - 10y + 25) = 1 \]

This simplifies to:

\[ (x - 2)^2 + (y - 5)^2 = 1 \]

Thus, the center \(C_1\) is \((2, 5)\) and the radius \(r_1 = 1\).

For the second circle, we rearrange the equation:

\[ x^2 + 4x + y^2 - 6y + 4 = 0 \]

Completing the square:

\[ (x^2 + 4x + 4) + (y^2 - 6y + 9) = 9 \]

This simplifies to:

\[ (x + 2)^2 + (y - 3)^2 = 9 \]

Thus, the center \(C_2\) is \((-2, 3)\) and the radius \(r_2 = 3\).

### Step 2: Calculate the distance between the centers

Using the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of the centers \(C_1(2, 5)\) and \(C_2(-2, 3)\):

\[ d = \sqrt{((-2) - 2)^2 + (3 - 5)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

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

Now we calculate the sum of the radii:

\[ r_1 + r_2 = 1 + 3 = 4 \]

### Step 4: Determine the number of common tangents

We compare the distance \(d\) with the sum of the radii:

- If \(d > r_1 + r_2\), there are 4 common tangents.
- If \(d = r_1 + r_2\), there are 3 common tangents.
- If \(d < r_1 + r_2\), there are 2 common tangents.

Here, we have:

\[ 2\sqrt{5} \approx 4.47 > 4 \]

Thus, since \(d > r_1 + r_2\), there are 4 common tangents.

### Conclusion

Therefore, the transverse common tangents of the circles are 4.

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