The two circles `x^(2)+y^(2)-2x-3=0` and `x^(2)+y^(2)-4x-6y-8=0` are such that

Options

A. they touch each other
B. they intersect each other
C. one lies inside the other
D. each lies outside the other

Answer

To determine the relationship between the two circles given by the equations \(x^2 + y^2 - 2x - 3 = 0\) and \(x^2 + y^2 - 4x - 6y - 8 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
1. **Circle 1**:
\[ x^2 + y^2 - 2x - 3 = 0 \]
Rearranging gives:
\[ (x^2 - 2x) + y^2 = 3 \]
Completing the square for \(x\):
\[ (x - 1)^2 - 1 + y^2 = 3 \implies (x - 1)^2 + y^2 = 4 \]
Thus, Circle 1 has center \(C_1(1, 0)\) and radius \(r_1 = 2\).

2. **Circle 2**:
\[ x^2 + y^2 - 4x - 6y - 8 = 0 \]
Rearranging gives:
\[ (x^2 - 4x) + (y^2 - 6y) = 8 \]
Completing the square for \(x\) and \(y\):
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 8 \implies (x - 2)^2 + (y - 3)^2 = 21 \]
Thus, Circle 2 has center \(C_2(2, 3)\) and radius \(r_2 = \sqrt{21}\).

### Step 2: Calculate the distance between the centers
The distance \(d\) between the centers \(C_1(1, 0)\) and \(C_2(2, 3)\) is given by:
\[ d = \sqrt{(2 - 1)^2 + (3 - 0)^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 3: Compare the distance with the sum and difference of the radii
1. **Sum of the radii**:
\[ r_1 + r_2 = 2 + \sqrt{21} \]
Approximating \(\sqrt{21} \approx 4.58\):
\[ r_1 + r_2 \approx 2 + 4.58 = 6.58 \]

2. **Difference of the radii**:
\[ r_2 - r_1 = \sqrt{21} - 2 \approx 4.58 - 2 = 2.58 \]

### Step 4: Analyze the relationship
- We have:
\[ d \approx \sqrt{10} \approx 3.16 \]
- Now we check the conditions:
- \(r_1 + r_2 > d\) (i.e., \(6.58 > 3.16\)) - True
- \(d > |r_2 - r_1|\) (i.e., \(3.16 > 2.58\)) - True

Since both conditions are satisfied, the circles intersect at two points.

### Conclusion
The two circles intersect each other at two points.

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