X and Y are centres of circles of a radii 9 cm and 2 cm respectively, XY = 17 cm. Z is the centre of a circle of radius r cm which touches of above circles externally. Given that `angle XZY = 90^(@)`, the value of r is

Options

A. 13 cm
B. 6 cm
C. 9 cm
D. 8 cm

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The radius of circle centered at X (R1) = 9 cm
- The radius of circle centered at Y (R2) = 2 cm
- The distance between centers X and Y (XY) = 17 cm
- The radius of the circle centered at Z (r) is what we need to find.
- The angle ∠XZY = 90°.

2. **Set Up the Relationships**:
- Since circle Z touches circle X externally, the distance XZ = R1 + r = 9 + r.
- Since circle Z touches circle Y externally, the distance YZ = R2 + r = 2 + r.

3. **Apply the Pythagorean Theorem**:
- According to the Pythagorean theorem, in triangle XZY:
\[ XY^2 = XZ^2 + YZ^2 \]
- Substituting the known values:
\[ 17^2 = (9 + r)^2 + (2 + r)^2 \]

4. **Calculate the Squares**:
- Calculate \(17^2\):
\[ 289 = (9 + r)^2 + (2 + r)^2 \]
- Expand the squares:
\[ (9 + r)^2 = 81 + 18r + r^2 \]
\[ (2 + r)^2 = 4 + 4r + r^2 \]
- Combine these:
\[ 289 = (81 + 18r + r^2) + (4 + 4r + r^2) \]
\[ 289 = 85 + 22r + 2r^2 \]

5. **Rearrange the Equation**:
- Move all terms to one side:
\[ 2r^2 + 22r + 85 - 289 = 0 \]
\[ 2r^2 + 22r - 204 = 0 \]

6. **Simplify the Quadratic Equation**:
- Divide the entire equation by 2:
\[ r^2 + 11r - 102 = 0 \]

7. **Factor the Quadratic**:
- Factor the quadratic equation:
\[ (r + 17)(r - 6) = 0 \]

8. **Find the Values of r**:
- Set each factor to zero:
\[ r + 17 = 0 \quad \Rightarrow \quad r = -17 \quad (\text{not valid since radius cannot be negative}) \]
\[ r - 6 = 0 \quad \Rightarrow \quad r = 6 \]

9. **Conclusion**:
- The radius \( r \) of the circle centered at Z is **6 cm**.