`x^2+y^2-10x-10y+41=0 and x^2+y^2-22x-10y+137=0`

Options

A. Meet at 1 point
B. Meet at 2 points
C. Does not meet
D. have same center

Answer

To solve the problem of finding the intersection points of the two circles given by the equations:

1. \( x^2 + y^2 - 10x - 10y + 41 = 0 \)
2. \( x^2 + y^2 - 22x - 10y + 137 = 0 \)

we will follow these steps:

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

The general form of a circle's equation is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center and \(r\) is the radius. We will complete the square for both equations.

**For the first circle:**

\[ x^2 - 10x + y^2 - 10y + 41 = 0 \]

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

- For \(x^2 - 10x\), we take \(-10/2 = -5\) and square it to get \(25\).
- For \(y^2 - 10y\), we take \(-10/2 = -5\) and square it to get \(25\).

Thus, we rewrite the equation:

\[ (x^2 - 10x + 25) + (y^2 - 10y + 25) = -41 + 25 + 25 \]

This simplifies to:

\[ (x - 5)^2 + (y - 5)^2 = 9 \]

So, the center \(C_1\) is \((5, 5)\) and the radius \(R_1 = 3\).

**For the second circle:**

\[ x^2 - 22x + y^2 - 10y + 137 = 0 \]

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

- For \(x^2 - 22x\), we take \(-22/2 = -11\) and square it to get \(121\).
- For \(y^2 - 10y\), we take \(-10/2 = -5\) and square it to get \(25\).

Thus, we rewrite the equation:

\[ (x^2 - 22x + 121) + (y^2 - 10y + 25) = -137 + 121 + 25 \]

This simplifies to:

\[ (x - 11)^2 + (y - 5)^2 = 9 \]

So, the center \(C_2\) is \((11, 5)\) and the radius \(R_2 = 3\).

### Step 2: Analyze the relationship between the circles

Now we have:

- Circle 1: Center \(C_1(5, 5)\), Radius \(R_1 = 3\)
- Circle 2: Center \(C_2(11, 5)\), Radius \(R_2 = 3\)

To find the distance \(d\) between the centers \(C_1\) and \(C_2\):

\[ d = \sqrt{(11 - 5)^2 + (5 - 5)^2} = \sqrt{6^2} = 6 \]

### Step 3: Determine the intersection condition

For two circles to intersect, the following conditions must hold:

1. They intersect at two points if \(d < R_1 + R_2\)
2. They are tangent (intersect at one point) if \(d = R_1 + R_2\)
3. They do not intersect if \(d > R_1 + R_2\)

Calculating \(R_1 + R_2\):

\[ R_1 + R_2 = 3 + 3 = 6 \]

Since \(d = 6\), we find that the circles are tangent to each other.

### Conclusion

The circles intersect at exactly one point.