A tangent PT is drawn to the circle `x^2+y^2=4` at the point `P(sqrt3,1)`. A straight line `L`, perpendicular to `PT` is a tangent to the circle `(x-3)^2+y^2=1` then find a common tangent of the two circles

Options

A. `x=4`
B. `y=2`
C. `x+sqrt(3) y =4`
D. `x+2 sqrt(2) y =6`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Find the slope of the tangent PT at point P(√3, 1) for the circle x² + y² = 4.

The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]

To find the slope of the tangent at point P, we differentiate the equation implicitly:
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(4) \implies 2x + 2y \frac{dy}{dx} = 0 \]
At point P(√3, 1):
\[ 2(\sqrt{3}) + 2(1) \frac{dy}{dx} = 0 \]
\[ 2\sqrt{3} + 2\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\sqrt{3} \]

### Step 2: Find the slope of the line L, which is perpendicular to PT.

The slope of PT is \(-\sqrt{3}\). The slope of line L, which is perpendicular to PT, is given by:
\[ m_1 = -\frac{1}{m} = -\frac{1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 3: Write the equation of line L.

Using the point-slope form of the line, we can write the equation of line L as:
\[ y - 1 = \frac{1}{\sqrt{3}}(x - \sqrt{3}) \]
Rearranging gives:
\[ y = \frac{1}{\sqrt{3}}x + \left(1 - \frac{\sqrt{3}}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}}x + 0 \]
Thus, the equation of line L can be rewritten as:
\[ \sqrt{3}y - x = 0 \]

### Step 4: Find the center and radius of the second circle.

The second circle is given by:
\[ (x - 3)^2 + y^2 = 1 \]
The center of this circle is (3, 0) and the radius is 1.

### Step 5: Find the perpendicular distance from the center (3, 0) to line L.

The formula for the perpendicular distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(\sqrt{3}y - x = 0\), we have:
- \(A = -1\)
- \(B = \sqrt{3}\)
- \(C = 0\)

Substituting the center (3, 0):
\[ d = \frac{|-1(3) + \sqrt{3}(0) + 0|}{\sqrt{(-1)^2 + (\sqrt{3})^2}} = \frac{3}{\sqrt{1 + 3}} = \frac{3}{2} \]

### Step 6: Set the distance equal to the radius of the second circle.

Since the distance \(d\) must equal the radius of the second circle (1):
\[ \frac{3}{2} = 1 \quad \text{(This is incorrect; we need to find the correct value of k)} \]
We need to adjust the equation of line L to find the correct tangent.

### Step 7: Adjust the equation of line L for the correct distance.

Let’s assume the equation of line L is:
\[ \sqrt{3}y - x + k = 0 \]
Then the distance from (3, 0) to this line is:
\[ d = \frac{|-\sqrt{3}(3) + 0 + k|}{\sqrt{(-1)^2 + (\sqrt{3})^2}} = \frac{|-3\sqrt{3} + k|}{2} \]
Setting this equal to the radius (1):
\[ \frac{|-3\sqrt{3} + k|}{2} = 1 \implies |-3\sqrt{3} + k| = 2 \]
This gives us two cases:
1. \(-3\sqrt{3} + k = 2 \implies k = 2 + 3\sqrt{3}\)
2. \(-3\sqrt{3} + k = -2 \implies k = -2 + 3\sqrt{3}\)

### Step 8: Write the equations of the common tangents.

Thus, the equations of the common tangents are:
1. \(\sqrt{3}y - x + (2 + 3\sqrt{3}) = 0\)
2. \(\sqrt{3}y - x + (-2 + 3\sqrt{3}) = 0\)

### Final Answer:

The common tangents of the two circles are:
1. \(\sqrt{3}y - x + (2 + 3\sqrt{3}) = 0\)
2. \(\sqrt{3}y - x + (-2 + 3\sqrt{3}) = 0\)