From an external point Q, the length of the tangents to a circle is 5 cm and the distance of Q from the centre is 8 cm. The radius of the circle is:

Options

A. 39 cm
B. 3 cm
C. `sqrt(39)` cm
D. 7 cm

Answer

To find the radius of the circle from the given information, we can use the Pythagorean theorem. Here's a step-by-step solution:

### Step 1: Understand the given information
We have:
- Length of the tangent from point Q to the circle (TQ) = 5 cm
- Distance from point Q to the center of the circle (OQ) = 8 cm
- We need to find the radius of the circle (OT).

### Step 2: Set up the triangle
We can form a right triangle OQT, where:
- O is the center of the circle,
- T is the point of tangency,
- Q is the external point.

In this triangle:
- OQ is the hypotenuse,
- OT is one leg (the radius),
- TQ is the other leg (the length of the tangent).

### Step 3: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ OQ^2 = OT^2 + TQ^2 \]

Substituting the known values:
\[ 8^2 = OT^2 + 5^2 \]

### Step 4: Calculate the squares
Calculating the squares:
- \( OQ^2 = 8^2 = 64 \)
- \( TQ^2 = 5^2 = 25 \)

### Step 5: Substitute and solve for OT
Now substituting the values back into the equation:
\[ 64 = OT^2 + 25 \]

To find \( OT^2 \), we rearrange the equation:
\[ OT^2 = 64 - 25 \]
\[ OT^2 = 39 \]

### Step 6: Find the radius
Taking the square root of both sides to find OT (the radius):
\[ OT = \sqrt{39} \]

Thus, the radius of the circle is \( \sqrt{39} \) cm.

### Final Answer
The radius of the circle is \( \sqrt{39} \) cm.

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