A circle of radius 2 has its centre at (2, 0) and another circle of radius 1 has its centre at (5, 0). A line is tangent to the two circles at point in the first quadrant. The y-intercept of the tangent line is

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `3sqrt(2)`
D. `4sqrt(2)`

Explanation

To find the y-intercept of the tangent line to the two circles, we will follow these steps:

### Step 1: Identify the equations of the circles
The first circle (C1) has a radius of 2 and is centered at (2, 0). Its equation is:
\[ (x - 2)^2 + y^2 = 2^2 \quad \text{(1)} \]
The second circle (C2) has a radius of 1 and is centered at (5, 0). Its equation is:
\[ (x - 5)^2 + y^2 = 1^2 \quad \text{(2)} \]

### Step 2: Set up the tangent line
Let the equation of the tangent line be in the slope-intercept form:
\[ y = mx + c \]
where \(m\) is the slope and \(c\) is the y-intercept.

### Step 3: Use the condition for tangency
For the line to be tangent to the first circle (C1), the distance from the center of the circle to the line must equal the radius. The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(y = mx + c\), we can rewrite it as:
\[ mx - y + c = 0 \]
Here, \(A = m\), \(B = -1\), and \(C = c\).

### Step 4: Apply the distance formula for the first circle
For circle C1 centered at (2, 0) with radius 2:
\[ \frac{|m(2) + (-1)(0) + c|}{\sqrt{m^2 + 1}} = 2 \]
This simplifies to:
\[ \frac{|2m + c|}{\sqrt{m^2 + 1}} = 2 \quad \text{(3)} \]

### Step 5: Apply the distance formula for the second circle
For circle C2 centered at (5, 0) with radius 1:
\[ \frac{|m(5) + (-1)(0) + c|}{\sqrt{m^2 + 1}} = 1 \]
This simplifies to:
\[ \frac{|5m + c|}{\sqrt{m^2 + 1}} = 1 \quad \text{(4)} \]

### Step 6: Solve equations (3) and (4)
From equation (3):
\[ |2m + c| = 2\sqrt{m^2 + 1} \quad \text{(5)} \]
From equation (4):
\[ |5m + c| = \sqrt{m^2 + 1} \quad \text{(6)} \]

### Step 7: Solve for \(c\)
We can express \(c\) from both equations:
From (5):
\[ c = 2\sqrt{m^2 + 1} - 2m \quad \text{(7)} \]
From (6):
\[ c = \sqrt{m^2 + 1} - 5m \quad \text{(8)} \]

### Step 8: Set equations (7) and (8) equal to each other
\[ 2\sqrt{m^2 + 1} - 2m = \sqrt{m^2 + 1} - 5m \]
Rearranging gives:
\[ \sqrt{m^2 + 1} = 3m \]
Squaring both sides:
\[ m^2 + 1 = 9m^2 \]
Thus:
\[ 8m^2 = 1 \quad \Rightarrow \quad m^2 = \frac{1}{8} \quad \Rightarrow \quad m = \frac{1}{2\sqrt{2}} \]

### Step 9: Substitute \(m\) back to find \(c\)
Using equation (7):
\[ c = 2\sqrt{\frac{1}{8} + 1} - 2\left(\frac{1}{2\sqrt{2}}\right) \]
Calculating:
\[ c = 2\sqrt{\frac{9}{8}} - \frac{1}{\sqrt{2}} = 2 \cdot \frac{3}{2\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer
The y-intercept of the tangent line is:
\[ c = 2\sqrt{2} \]