P and Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm which touches the above circles externally. Given that `anglePRQ = 90^@` , write an equation in x and solve it.

Answer

To solve the problem step by step, we will follow the given information and apply the Pythagorean theorem.

### Step 1: Understand the given information
We have:
- Circle with center P has a radius of 9 cm.
- Circle with center Q has a radius of 2 cm.
- Distance PQ = 17 cm.
- R is the center of a circle with radius x cm that touches both circles externally.
- Angle PRQ = 90°.

### Step 2: Set up the triangle PQR
In triangle PQR, we know that:
- PR = radius of circle at P + radius of circle at R = 9 + x.
- RQ = radius of circle at Q + radius of circle at R = 2 + x.
- PQ = 17 cm.

### Step 3: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ PQ^2 = PR^2 + RQ^2 \]

Substituting the values we have:
\[ 17^2 = (9 + x)^2 + (2 + x)^2 \]

### Step 4: Calculate \( PQ^2 \)
Calculating \( 17^2 \):
\[ 17^2 = 289 \]

### Step 5: Expand \( PR^2 \) and \( RQ^2 \)
Now we expand \( (9 + x)^2 \) and \( (2 + x)^2 \):
\[ (9 + x)^2 = 81 + 18x + x^2 \]
\[ (2 + x)^2 = 4 + 4x + x^2 \]

### Step 6: Combine the equations
Now substituting back into the equation:
\[ 289 = (81 + 18x + x^2) + (4 + 4x + x^2) \]
Combine like terms:
\[ 289 = 81 + 4 + 18x + 4x + 2x^2 \]
\[ 289 = 85 + 22x + 2x^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 2x^2 + 22x + 85 - 289 = 0 \]
\[ 2x^2 + 22x - 204 = 0 \]

### Step 8: Simplify the equation
Dividing the entire equation by 2:
\[ x^2 + 11x - 102 = 0 \]

### Step 9: Factor the quadratic equation
Now we need to factor the quadratic equation:
We are looking for two numbers that multiply to -102 and add to 11. The numbers are 17 and -6.
So we can write:
\[ (x + 17)(x - 6) = 0 \]

### Step 10: Solve for x
Setting each factor to zero gives us:
1. \( x + 17 = 0 \) → \( x = -17 \) (not valid since radius cannot be negative)
2. \( x - 6 = 0 \) → \( x = 6 \)

Thus, the radius \( x \) of the circle centered at R is **6 cm**.

### Final Answer
The value of \( x \) is **6 cm**.

---