The circles `x^(2)+y^(2)-2x-4y-20=0, x^(2)+y^(2)+4x-2y+4=0` are

Options

A. one lies out side the other
B. one lies completely inside the other
C. touch externally
D. touch internally

Answer

To determine the relationship between the two circles given by the equations \( S_1: x^2 + y^2 - 2x - 4y - 20 = 0 \) and \( S_2: x^2 + y^2 + 4x - 2y + 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the equations in standard form
We start by rewriting the equations of the circles in standard form, which is \( (x - h)^2 + (y - k)^2 = r^2 \).

**For Circle \( S_1 \):**
\[ x^2 + y^2 - 2x - 4y - 20 = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 - 4y = 20 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 20 \]
\[ (x - 1)^2 + (y - 2)^2 = 25 \]
Thus, the center \( C_1 \) is \( (1, 2) \) and the radius \( R_1 = 5 \).

**For Circle \( S_2 \):**
\[ x^2 + y^2 + 4x - 2y + 4 = 0 \]
Rearranging gives:
\[ x^2 + 4x + y^2 - 2y = -4 \]
Completing the square:
\[ (x + 2)^2 - 4 + (y - 1)^2 - 1 = -4 \]
\[ (x + 2)^2 + (y - 1)^2 = 1 \]
Thus, the center \( C_2 \) is \( (-2, 1) \) and the radius \( R_2 = 1 \).

### Step 2: Calculate the distance between the centers
Now, we find the distance \( d \) between the centers \( C_1 \) and \( C_2 \):
\[ d = \sqrt{(1 - (-2))^2 + (2 - 1)^2} = \sqrt{(1 + 2)^2 + (2 - 1)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 3: Compare the distance with the sum of the radii
Next, we compare the distance \( d \) with the sum of the radii \( R_1 + R_2 \):
\[ R_1 + R_2 = 5 + 1 = 6 \]
Since \( d = \sqrt{10} \) and \( R_1 + R_2 = 6 \), we note that:
\[ \sqrt{10} < 6 \]

### Step 4: Determine the relationship between the circles
Since the distance between the centers \( d \) is less than the sum of the radii \( R_1 + R_2 \), it indicates that one circle lies completely inside the other.

### Final Conclusion
Thus, the circles \( S_1 \) and \( S_2 \) are such that one lies completely inside the other.