If two circles `(x-1)^(2)+(y-3)^(2)=r^(2)` and `x^(2)+y^(2)-8x+2y+8=0` intersect in two distinct points , then

Options

A. `2 ltr lt 8`
B. `r lt 2`
C. `r =2`
D. `r gt 2 `

Answer

To solve the problem, we need to analyze the two circles given and determine the conditions under which they intersect at two distinct points.

### Step 1: Identify the centers and radii of the circles

1. **First Circle**: The equation is \((x-1)^2 + (y-3)^2 = r^2\).
- Center \(C_1 = (1, 3)\)
- Radius \(r_1 = r\)

2. **Second Circle**: The equation is \(x^2 + y^2 - 8x + 2y + 8 = 0\).
- We can rewrite this in standard form by completing the square.

Rearranging gives:
\[ (x^2 - 8x) + (y^2 + 2y) + 8 = 0 \]
Completing the square for \(x\):
\[ x^2 - 8x = (x-4)^2 - 16 \]
Completing the square for \(y\):
\[ y^2 + 2y = (y+1)^2 - 1 \]
Substituting back, we have:
\[ (x-4)^2 - 16 + (y+1)^2 - 1 + 8 = 0 \]
Simplifying this gives:
\[ (x-4)^2 + (y+1)^2 - 9 = 0 \]
Thus, we can express it as:
\[ (x-4)^2 + (y+1)^2 = 3^2 \]
- Center \(C_2 = (4, -1)\)
- Radius \(r_2 = 3\)

### Step 2: Calculate the distance between the centers

The distance \(d\) between the centers \(C_1\) and \(C_2\) is given by:
\[ d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Establish the condition for intersection

For two circles to intersect at two distinct points, the following condition must hold:
\[ d < r_1 + r_2 \]
Substituting the known values:
\[ 5 < r + 3 \]

### Step 4: Solve the inequality

Rearranging the inequality gives:
\[ r > 5 - 3 \]
\[ r > 2 \]

### Conclusion

Thus, the condition for the circles to intersect at two distinct points is:
\[ r > 2 \]

### Final Answer

The correct option is that \(r\) must be greater than 2.

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