**(a) $ \boldsymbol{ r \ = \ 2 \pi C } $**

**(b) $ \boldsymbol{ r \ = \ \dfrac{ C – \pi }{ 2 } } $**

**(c) $ \boldsymbol{ r \ = \ \dfrac{ C }{ 2 \pi } } $**

**(d) $ \boldsymbol{ r \ = \ C – 2 \pi } $**

This question aims to develop an understanding of the **algebraic simplification** of the equation for the **circumference of a circle** using basic **arithmetic operations**.

The **circumference of a circle** is the **length of its outer periphery**. It is mathematically defined by the following **formula**:

**\[ \boldsymbol{ C \ = \ 2 \pi r } \]**

Where $ C $ represents the **circumference** and $ r $ represents the **radius** of the subject circle. Now this **formula can be directly used** to calculate the circumference **given the radius** of the circle, however, if we were **to evaluate** the value of $ r $ **given the circumference**, then we may have to **modify** it a little bit. This **rearrangement** process is called the **algebraic simplification** process which is further explained in the following solution.

## Expert Answer

Given the **formula of the circumference** of the circle:

\[ C \ = \ 2 \pi r \]

**Dividing both sides by $ 2 $:**

\[ \dfrac{ C }{ 2 } \ = \ \dfrac{ 2 \pi r }{ 2 } \]

\[ \Rightarrow \dfrac{ C }{ 2 } \ = \Â \pi r \]

**Dividing both sides by $ \pi $:**

\[ \dfrac{ C }{ 2 \pi } \ = \ \dfrac{ \pi r }{ \pi } \]

\[ \Rightarrow \dfrac{ C }{ 2 \pi } \ = \Â r \]

**Exchanging sides:**

\[ r \ = \ \dfrac{ C }{ 2 \pi } \]

Which is the required expression. If we **compare it** with the given options, we can see that **option (c) is the right answer**.

## Numerical Result

\[ r \ = \ \dfrac{ C }{ 2 \pi } \]

## Example

The **area of a circle** is given by the following formula:

\[ A \ = \ \pi r^{ 2 } \]

**Find the value of $ r $.**

**Dividing the above equation by $ \pi $:**

\[ \dfrac{ A }{ \pi } \ = \ \dfrac{ \pi r^{ 2 } }{ \pi } \]

\[ \Rightarrow \dfrac{ A }{ \pi } \ = \ r^{ 2 } \]

Taking **square root** on both sides:

\[ \sqrt{ \dfrac{ A }{ \pi } } \ = \ \sqrt{ r^{ 2 } } \]

Since $ \sqrt{ r^{ 2 } } \ = \ \pm r $, above equation becomes:

\[ \sqrt{ \dfrac{ A }{ \pi } } \ = \ \pm r \]

**Exchanging sides:**

\[ r \ = \ \pm \sqrt{ \dfrac{ A }{ \pi } } \]