From an external point Q, the length of tangent to a circle is 12 cm and the distance of Q from the centre of circle is 13 cm. The radius of circle (in cm) is

Options

A. 10
B. 5
C. 12
D. 7

Answer

To solve the problem, we will use the Pythagorean theorem. Here are the steps:

### Step 1: Identify the elements of the problem
Let:
- \( O \) be the center of the circle.
- \( Q \) be the external point from which the tangent is drawn.
- \( P \) be the point of tangency on the circle.

We know:
- Length of the tangent \( PQ = 12 \) cm
- Distance from point \( Q \) to the center \( O \) is \( OQ = 13 \) cm
- We need to find the radius of the circle, which is \( OP \).

### Step 2: Use the Pythagorean theorem
In the right triangle \( OQP \):
- \( OQ^2 = OP^2 + PQ^2 \)

Substituting the known values:
- \( 13^2 = OP^2 + 12^2 \)

### Step 3: Calculate the squares
Calculate \( 13^2 \) and \( 12^2 \):
- \( 13^2 = 169 \)
- \( 12^2 = 144 \)

### Step 4: Set up the equation
Now we can set up the equation:
- \( 169 = OP^2 + 144 \)

### Step 5: Solve for \( OP^2 \)
Rearranging the equation gives:
- \( OP^2 = 169 - 144 \)
- \( OP^2 = 25 \)

### Step 6: Find \( OP \)
Taking the square root of both sides:
- \( OP = \sqrt{25} = 5 \) cm

Thus, the radius of the circle is \( 5 \) cm.

### Final Answer:
The radius of the circle is \( 5 \) cm.

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