This problem aims to find the cartesian equation for the curve and after that identify the curve. To better understand the problem, you should be familiar with **cartesian coordinate systems, polar coordinates, **and **conversion** from **polar** to **cartesian coordinates.**

A **two-dimensional coordinate system** in which a **point** on a plane is determined by a **distance** from a **pole** (reference point) and an **angle** from the **reference plane,** is known as the **polar coordinate.** On the other hand,** spherical coordinates** are the **3 coordinates** that determine the location of a **point** in a **3-dimensional** trajectory. We can convert **cartesian coordinates** to **polar coordinates** using the equations:

\[ x = r\cos\theta \]

\[ y = r\sin\theta \]

Where $r$ is the **distance** from the **reference point,** and can be found using $r = \sqrt{x^2 + y^2}$,

and $\theta$ is the **angle** with the **plane,** which can be **calculated** as $\theta = \tan^{-1}{\dfrac{y}{x}}$.

## Expert Answer

We know that $r$ and $\theta$ are called **polar coordinates** of $P$ such that $P(r,\theta).

Now we are given a **polar equation** of the **curve** that is:

\[ r = 5\cos\theta \]

To **convert** the above **equation** into the form of $x^2 + y^2 = r^2$, we will be **multiplying** both **sides** by $r$:

\[ r^2 = 5r\cos\theta \]

First, we will **transform** the above **polar equation** from **polar** to **cartesian coordinates.**

**Transformation** of **polar** to **Cartesian coordinates** can be done using the concept,

\[x^2 + y^2 = r^2, \space x = r\cos\theta \]

Therefore, the given curve in the **cartesian coordinates** can be written as:

\[ x^2 + y^2 = 5x \]

Rewriting the **equation** as:

\[ x^2 + y^2 – 5x = 0 \]

Applying the **technique** for **completing** the **square:**

\[ x^2 + y^2 – 5x + \dfrac{25}{4} – \dfrac{25}{4} = 0 \]

\[ (x – \dfrac{5}{2})^2 + y^2 = \dfrac{25}{4} \]

This **equation** denotes a **circle** that is **centered** at a **point** $(\dfrac{5}{2},0)$ with **radius** $\dfrac{5}{2}$.

## Numerical Result

The **polar equation** $r = 5 \cos \theta$ **transformed** into **cartesian coordinates** as $(x – \dfrac{5}{2})^2 + y^2 = \dfrac{25}{4}$, which represents a **circle** with **center point** $(\dfrac{5}{2},0)$ and **radius** $\dfrac{5}{2}$.

## Example

Identify the **curve** by figuring out the **cartesian equation** for $r^2 \cos2 \theta = 1$.

We know that $r$ and $\theta$ are **polar coordinates** of $P$, such that $P(r,\theta).

We are given a **polar equation** of the **curve** that is:

\[r^2 \cos2 \theta = 1\]

First, we will **transform** the above **polar equation** from **polar** to **cartesian coordinates.**

**Transformation** of **polar** to **Cartesian coordinates** can be done using the concept,

\[x^2 + y^2 = r^2, \space x = r\cos\theta, \space y = r\sin\theta \]

**Therefore,**

\[r^2\cos2\theta = 1\]

Using the **trigonometric formula** for $\cos2\theta$, that is:

\[ \cos2\theta = \cos^2\theta – \sin^2\theta \]

**Rewriting** the equation as:

\[r^2(\cos^2\theta – \sin^2\theta) = 1\]

\[r^2\cos^2\theta – r^2\sin^2\theta = 1\]

\[(r\cos\theta)^2 – (r\sin\theta)^2 = 1\]

**Plugging** the values of $ x = r\cos\theta, \space y = r\sin\theta $ gives:

\[ x^2 + y^2 = 1 \]

Therefore, the **cartesian equation** $ x^2 + y^2 = 1$ represents a **hyperbola.**