The radical centre of the three circles is at the origin. The equations of the two of the circles are `x^(2)+y^(2)=1 and x^(2)+y^(2)+4x+4y-1=0`. If the third circle passes through the points (1, 1) and (-2, 1), and its radius can be expressed in the form of `(p)/(q)`, where p and q are relatively prime positive integers. Find the value of `(p+q)`.

Answer

To solve the given problem step by step, we will follow the outlined approach based on the information provided.

### Step 1: Understand the Given Circles
We have two circles:
1. Circle 1: \( x^2 + y^2 = 1 \)
2. Circle 2: \( x^2 + y^2 + 4x + 4y - 1 = 0 \)

We can rewrite Circle 2 in standard form:
\[ x^2 + y^2 + 4x + 4y = 1 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 + 4x) + (y^2 + 4y) = 1 \]
\[ (x + 2)^2 - 4 + (y + 2)^2 - 4 = 1 \]
\[ (x + 2)^2 + (y + 2)^2 = 9 \]
This means Circle 2 has center \((-2, -2)\) and radius \(3\).

### Step 2: Find the Equation of the Third Circle
The third circle passes through the points \((1, 1)\) and \((-2, 1)\). The general equation of a circle passing through two points can be expressed as:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) + \lambda \cdot \text{determinant} = 0 \]
Where \((x_1, y_1) = (1, 1)\) and \((x_2, y_2) = (-2, 1)\).

### Step 3: Set Up the Equation
Using the points:
\[ (x - 1)(x + 2) + (y - 1)(y - 1) + \lambda \cdot \text{det} = 0 \]
The determinant is:
\[ \begin{vmatrix} x & y & 1 \\ 1 & 1 & 1 \\ -2 & 1 & 1 \end{vmatrix} = x(1 - 1) - y(1 + 2) + 1(1 + 2) = -3y + 3 \]
Thus, the equation becomes:
\[ (x - 1)(x + 2) + (y - 1)^2 + \lambda(-3y + 3) = 0 \]

### Step 4: Find the Radical Axis
The radical axis of the first two circles is given by:
\[ s_1 - s_2 = 0 \]
Where \(s_1\) and \(s_2\) are the equations of the circles. We can express this as:
\[ x^2 + y^2 - 1 - (x^2 + y^2 + 4x + 4y - 1) = 0 \]
This simplifies to:
\[ -4x - 4y + 2 = 0 \quad \Rightarrow \quad x + y = \frac{1}{2} \]

### Step 5: Substitute into the Radical Axis
Since the radical axis passes through the origin, substituting \(x = 0\) and \(y = 0\) gives us:
\[ 0 + 0 = \frac{1}{2} \quad \Rightarrow \quad \lambda = 0 \]

### Step 6: Finalize the Equation of the Third Circle
Substituting \(\lambda = 0\) into the equation:
\[ (x - 1)(x + 2) + (y - 1)^2 = 0 \]
Expanding this:
\[ x^2 + x - 2 + (y - 1)^2 = 0 \]
\[ x^2 + x + y^2 - 2y - 2 = 0 \]

### Step 7: Find the Radius
Rearranging gives:
\[ x^2 + y^2 + x - 2y - 2 = 0 \]
Completing the square:
\[ (x + \frac{1}{2})^2 + (y - 1)^2 = \frac{9}{4} \]
Thus, the radius \(r = \frac{3}{2}\).

### Step 8: Express in the Form \( \frac{p}{q} \)
Here, \(p = 3\) and \(q = 2\). Since \(3\) and \(2\) are relatively prime, we find:
\[ p + q = 3 + 2 = 5 \]

### Final Answer
The value of \(p + q\) is \(5\).