P and Q are centres of two circles with radii 9 cm and 2 cm respectively, where PQ = 17 cm, R is the centre of another circle of radius x cm which touches each of the above two circles externally. If `angle PRQ = 90^(@)`, then the value of x is -

Options

A. 4 cm
B. 6 cm
C. 7 cm
D. 8 cm

Answer

To solve the problem step by step, we need to find the radius \( x \) of the circle centered at point \( R \) that touches the two given circles with centers \( P \) and \( Q \) externally. Here’s how we can approach this:

### Step 1: Understand the Configuration
We have two circles:
- Circle with center \( P \) and radius \( 9 \) cm.
- Circle with center \( Q \) and radius \( 2 \) cm.
- The distance \( PQ = 17 \) cm.
- We need to find the radius \( x \) of the circle centered at \( R \) that touches both circles externally.

### Step 2: Set Up the Equation
Since the circles touch each other externally, we can use the following relationship based on the distances:
- The distance from \( P \) to \( R \) is \( PR = 9 + x \) (since they touch externally).
- The distance from \( Q \) to \( R \) is \( QR = 2 + x \).

### Step 3: Apply the Pythagorean Theorem
Given that \( \angle PRQ = 90^\circ \), we can apply the Pythagorean theorem:
\[ PQ^2 = PR^2 + QR^2 \]
Substituting the known values:
\[ 17^2 = (9 + x)^2 + (2 + x)^2 \]

### Step 4: Expand the Equation
Now, we expand both sides:
\[ 289 = (9 + x)^2 + (2 + x)^2 \]
Expanding the squares:
\[ (9 + x)^2 = 81 + 18x + x^2 \]
\[ (2 + x)^2 = 4 + 4x + x^2 \]
Combining these:
\[ 289 = (81 + 18x + x^2) + (4 + 4x + x^2) \]
\[ 289 = 81 + 4 + 18x + 4x + 2x^2 \]
\[ 289 = 85 + 22x + 2x^2 \]

### Step 5: Rearrange the Equation
Now, rearranging gives us:
\[ 2x^2 + 22x + 85 - 289 = 0 \]
\[ 2x^2 + 22x - 204 = 0 \]

### Step 6: Simplify the Quadratic Equation
Dividing the entire equation by 2:
\[ x^2 + 11x - 102 = 0 \]

### Step 7: Factor or Use the Quadratic Formula
Now, we can either factor or use the quadratic formula. Let's use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 11, c = -102 \):
\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot (-102)}}{2 \cdot 1} \]
\[ x = \frac{-11 \pm \sqrt{121 + 408}}{2} \]
\[ x = \frac{-11 \pm \sqrt{529}}{2} \]
\[ x = \frac{-11 \pm 23}{2} \]

### Step 8: Calculate Possible Values for \( x \)
Calculating the two possible values:
1. \( x = \frac{12}{2} = 6 \)
2. \( x = \frac{-34}{2} = -17 \) (not valid since radius cannot be negative)

Thus, the only valid solution is:
\[ x = 6 \text{ cm} \]

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

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