Two circles C(O, r) and C'(O', r') touch externally at P(3, 1). If the co-ordinates of O and O' are (1, p) and (q, -2) repectively. Their areas are in the ratio 4 : 9. Find the value of `p^(2)+q^(2)`.

Answer

To solve the problem, we need to find the value of \( p^2 + q^2 \) given the conditions of the circles. Let's break it down step by step.

### Step 1: Understand the given information
We have two circles \( C(O, r) \) and \( C'(O', r') \) that touch externally at the point \( P(3, 1) \). The centers of the circles are given as \( O(1, p) \) and \( O'(q, -2) \). The areas of the circles are in the ratio \( 4:9 \).

### Step 2: Use the area ratio to find a relationship between the radii
The area of a circle is given by \( \pi r^2 \). Therefore, the ratio of the areas of the two circles can be expressed as:
\[ \frac{\pi r^2}{\pi r'^2} = \frac{4}{9} \]
This simplifies to:
\[ \frac{r^2}{r'^2} = \frac{4}{9} \]
Taking square roots gives:
\[ \frac{r}{r'} = \frac{2}{3} \]
Thus, we can express \( r \) in terms of \( r' \):
\[ r = \frac{2}{3} r' \]

### Step 3: Use the distance formula to express the radii
The distance \( OP \) (which is the radius \( r \)) can be calculated using the distance formula:
\[ r = \sqrt{(1 - 3)^2 + (p - 1)^2} = \sqrt{(-2)^2 + (p - 1)^2} = \sqrt{4 + (p - 1)^2} \]

Similarly, the distance \( O'P \) (which is the radius \( r' \)) is:
\[ r' = \sqrt{(q - 3)^2 + (-2 - 1)^2} = \sqrt{(q - 3)^2 + (-3)^2} = \sqrt{(q - 3)^2 + 9} \]

### Step 4: Set up the equation using the ratio of the radii
From the relationship \( r = \frac{2}{3} r' \), we can substitute the expressions for \( r \) and \( r' \):
\[ \sqrt{4 + (p - 1)^2} = \frac{2}{3} \sqrt{(q - 3)^2 + 9} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ 4 + (p - 1)^2 = \frac{4}{9} \left((q - 3)^2 + 9\right) \]
Multiplying through by 9 to eliminate the fraction:
\[ 9(4 + (p - 1)^2) = 4((q - 3)^2 + 9) \]
Expanding both sides:
\[ 36 + 9(p - 1)^2 = 4(q - 3)^2 + 36 \]
Subtracting 36 from both sides:
\[ 9(p - 1)^2 = 4(q - 3)^2 \]

### Step 6: Rearranging the equation
Dividing both sides by 9:
\[ (p - 1)^2 = \frac{4}{9}(q - 3)^2 \]
Taking the square root gives:
\[ \frac{p - 1}{q - 3} = \frac{2}{3} \quad \text{or} \quad \frac{p - 1}{q - 3} = -\frac{2}{3} \]

### Step 7: Solving the equations
1. From \( \frac{p - 1}{q - 3} = \frac{2}{3} \):
\[ 3(p - 1) = 2(q - 3) \implies 3p - 3 = 2q - 6 \implies 3p - 2q = -3 \quad \text{(Equation 1)} \]

2. From \( \frac{p - 1}{q - 3} = -\frac{2}{3} \):
\[ 3(p - 1) = -2(q - 3) \implies 3p - 3 = -2q + 6 \implies 3p + 2q = 9 \quad \text{(Equation 2)} \]

### Step 8: Solve the system of equations
Using Equation 1 and Equation 2:
1. \( 3p - 2q = -3 \)
2. \( 3p + 2q = 9 \)

Adding these two equations:
\[ (3p - 2q) + (3p + 2q) = -3 + 9 \implies 6p = 6 \implies p = 1 \]

Substituting \( p = 1 \) into Equation 1:
\[ 3(1) - 2q = -3 \implies 3 - 2q = -3 \implies -2q = -6 \implies q = 3 \]

### Step 9: Calculate \( p^2 + q^2 \)
Now that we have \( p = 1 \) and \( q = 3 \):
\[ p^2 + q^2 = 1^2 + 3^2 = 1 + 9 = 10 \]

### Final Answer
Thus, the value of \( p^2 + q^2 \) is \( \boxed{10} \).