AB is a chord of a circle with O as centre. C is a point on the circle such that OC is perpendicular to AB and OC intersects AB at P. If PC = 2 cm and AB = 6 cm then the diameter of the circle is

Options

A. 6 cm
B. 6.5 cm
C. 13 cm
D. 12 cm

Answer

To solve the problem, we will use the properties of circles and right triangles. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
We have a circle with center O, a chord AB, and a perpendicular line OC from the center O to the chord AB, intersecting it at point P. We know that PC = 2 cm and AB = 6 cm.

### Step 2: Determine the Lengths
Since AB is a chord of length 6 cm, we can find the lengths of PA and PB. Because OC is perpendicular to AB, it bisects the chord. Therefore:
- PA = PB = AB / 2 = 6 cm / 2 = 3 cm.

### Step 3: Use the Right Triangle
In triangle OPC, we can apply the Pythagorean theorem. We have:
- OP = r - PC, where r is the radius of the circle.
- PC = 2 cm.
- PA = 3 cm.

Using the Pythagorean theorem:
\[ OP^2 + PC^2 = PA^2 \]
Substituting the known values:
\[ OP^2 + 2^2 = 3^2 \]

### Step 4: Substitute and Solve
Now, substituting the values:
\[ OP^2 + 4 = 9 \]
\[ OP^2 = 9 - 4 \]
\[ OP^2 = 5 \]
\[ OP = \sqrt{5} \]

### Step 5: Relate OP to the Radius
Since OP = r - PC, we can write:
\[ \sqrt{5} = r - 2 \]
Thus, solving for r:
\[ r = \sqrt{5} + 2 \]

### Step 6: Calculate the Diameter
The diameter D of the circle is twice the radius:
\[ D = 2r = 2(\sqrt{5} + 2) \]
\[ D = 2\sqrt{5} + 4 \]

### Step 7: Approximate the Diameter
To find a numerical value, we can approximate:
\[ \sqrt{5} \approx 2.236 \]
So:
\[ D \approx 2(2.236) + 4 \]
\[ D \approx 4.472 + 4 \]
\[ D \approx 8.472 \]

### Conclusion
Thus, the diameter of the circle is approximately \( 8.472 \) cm.