Circle and its properties

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# Circle and its properties
## Analytic Geometry
### Academy of Mathematics
### High School UPAEP Angelópolis
### Fall 2022

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# Geometrical elements of a circle

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# First approach

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# Elements of a circle

.pull-left[
 We only need **2** elements to construct a circle:

- the center `$\color{red}{C(h,k)}$`: it is a point in the plane

- the radius `$\color{red}{r}$`: it is the distance between the center and a point on the circle
 
 * Can `$\color{red}{r}$` be negative?
 
 * What happen if `$\color{red}{r}=0$`?

]

.pull-right[

<img src="circle.png" width="110%" style="display: block; margin: auto;" />
]

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PART I

Standard equation of the circle

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# Circle in standard form

.pull-left[


The **STANDARD EQUATION** of a circle with center at `$\color{red}{C}(\color{green}{h},\color{purple}{k})$` and radius `$\color{blue}{r}$` is

$$
 (x-\color{green}{h})^2 + (y-\color{purple}{k})^2 = \color{blue}{r}^2
$$
]

.pull-right[
<img src="circle-centered-(h,k).png" width="120%" style="display: block; margin: auto;" />
]

---

# Example 1 (1/2)

**Problem 1**

Find the equation of a circle centered at the origin with radius `$6$`. Sketch the graph of this circle.

**Solution**

We identify the elements of the circle:

$$
`\begin{gathered}
\text{center: }& C(\color{green}{0},\color{purple}{0}), \\
\text{radius: }& r=\color{blue}{6}
\end{gathered}`
$$
We substitute into the standard equation and reduce:

$$
`\begin{gathered}
(x-\color{green}{0})^2 + (y-\color{purple}{0})^2 = (\color{blue}{6})^2 \\
\\
x^2+ y^2 = \color{blue}{36}
\end{gathered}`
$$

---
# Example 1(2/2)

The circle represented by the equation `$x^2 + y^2 = \color{blue}{36}$` is given in the figure.

---

# Example 2 (1/2)

.pull-left[
 **Problem 2**

Find the standard equation of the circle in the figure.

<img src="circle-e2.png" width="90%" style="display: block; margin: auto;" />
]

.pull-right[
 **Solution**

We note that `$\color{red}{C}(\color{green}{-5}, \color{purple}{0})$`. On the other hand, the point `$P(-2, 0)$` lies on the circle, so the radius is equal to the distance between `$C$` and `$P$`, this is

$$
 \begin{aligned}
 \color{blue}{r} &= d(C, P) \\\
 & = \sqrt{(\color{green}{-5} - (-2))^{2} + (0-0)^2} \\\
 & = \sqrt{(-5+2)^2 + 0^2} \\\
 & = \sqrt{(-3)^2 + 0} \\\
 & = \sqrt{9} \\\
 & = 3
 \end{aligned}
$$

]

---

# Example 2(2/2)

Then, the center is `$\color{red}{C}(\color{green}{-5}, \color{purple}{0})$` and the radius is `$\color{blue}{r = 3}$`.

Next, we substitute into the standard equation and reduce terms:

$$
`\begin{gathered}
(x-(\color{green}{-5}))^2 + (y - \color{purple}{0})^2 = (\color{blue}{3})^2 \\
 \\
 (x + \color{green}{5})^2 + y^2 = \color{blue}{9}
\end{gathered}`
$$

In conclusion, the requested equation is

$$
 (x + \color{green}{5})^2 + y^2 = \color{blue}{9}
$$

---
class: inverse, center

PART II

General equation of the circle

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# From the standard equation to the general equation

Given a circle centered at `$(\color{green}{h}, \color{purple}{k})$` with radius `$\color{blue}{r}$`, to find the **general equation** of the circle, we have to follow the next 3 steps:

- **Step 1**: Find the standard equation of the circle.

- **Step 2**: Expand the binomials that appear.

- **Step 3**: Reorder the terms and reduce similar terms.

---
# Example 3(1/2)

.pull-left[
 **Problem 3**

Find the general equation of the circle centered at `$C(\color{green}{2}, \color{purple}{-3})$` with radius `$\color{blue}{5}$`.

**Solution**

We note that our data is the following:

$$
`\begin{gathered}
\text{center: }& C(\color{green}{2},\color{purple}{-3}), \\
\text{radius: }& r=\color{blue}{5}
\end{gathered}`
$$
]

.pull-right[
 **Step 1**: The standard equation of the circle is given by

$$
(x- \color{green}{2})^2 + (y - (\color{purple}{-3}))^2 = \color{blue}{5}^2
$$
this is
 $$
(x- \color{green}{2})^2 + (y + \color{purple}{3})^2 = \color{blue}{25}
$$

**Step 2**: Now, we expand the binomials to get

$$
(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25
$$
]

---
# Example 3 (2/2)

**Step 3**: Now, let's reorder and reduce the terms. We procced as follows:

$$
`\begin{gathered}
(x^2 - 4x + 4) + (y + 6y + 9) = 25 \\
\\
x^2 + y^2 - 4x +6y + 4 + 9 - 25 = 0 \\
\\
x^2 + y^2 - 4x + 6y - 12 = 0
\end{gathered}`
$$
Then, the general equation of the circle centered at `$C(2, -3)$` with radius `$5$` is

$$
\color{red}{x^2 + y^2 - 4x + 6y - 12 = 0}
$$

---
class: inverse, center

PART III

Getting the center and the radius from an equation

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#Center and radius from the standard equation

.pull-left[
 In this case, we begin with an equation of the form

$$(x+5)^2 + y^2 = 81
$$

So, because the equation looks like the _standard equation of a circle_, we **claim** that we just have to look for the center and the radius of a circle.
]

.pull-right[
 In this case, to find the center we just have to:

- For the **center**: take the opposite sign of the numbers that appear next to the variables `$x$` and `$y$`, respectively.
 - For the **radius**: take the square root of the number on the right hand.

]

---
# Example 4

**Problem 4**

Find the elements of the figure represented by the following equation

$$ (x\color{green}{+5})^2 + y^2 = \color{blue}{81}
$$

**Solution**

We note that it is a standard equation of a circle. In this case, the center is `$C(\color{green}{-5}, \color{purple}{0})$`: the second coordinate is zero because no number appears next to `$y$`, so, we actually have `$(y-\color{purple}{0})^2$`. On the other hand, the radius is `$r = \color{blue}{\sqrt{81}} = \color{blue}{9}$`. Then, the final answer is

$$
`\begin{gathered}
\text{center: }& C(\color{green}{-5},\color{purple}{0}), \\
\text{radius: }& r=\color{blue}{9}
\end{gathered}`
$$

---
# Example 5

**Problem 5**

Find the elements of the figure represented by the following equation

$$ (x\color{green}{-9})^2 + (y \color{purple}{+7})^2 = \color{blue}{51}
$$

**Solution**

We note that it is a standard equation of a circle. Then, the final answer is

$$
`\begin{gathered}
\text{center: }& C(\color{green}{9},\color{purple}{-7}), \\
\text{radius: }& r=\color{blue}{\sqrt{51}}
\end{gathered}`
$$

---
# From the general equation to the standard equation

Consider the equation given by

$$\color{darkred}{A}x^2 + \color{darkblue}{B}y^2+Cx+Dy+F=0
$$

If `$\color{darkred}{A} = \color{darkblue}{B}$`, then we _expect_ a **general equation of a circle**. In this case, the strategy is **transforming into the standard equation** by _completing squares_ for both `$x$` and `$y$`: later we just apply the strategy in the case of the standard equation.

---
# Example 6 (1/3)

**Problem 6**

Find the geometrical elements of the figure with equation given by

$$
x^2 + y^2 - 6x + 8y +9 = 0
$$

**Solution**

We note that the coefficient of `$x^2$` and `$y^2$` is equal to `$1$`, then we can _expect_ that it is the **general equation of a circle**. Then, the _geometrical elements_ that we are looking for are the **center** and the **radius** of the circle.

---
# Example 6 (2/3)

**Step 1**: Group the terms with the same variable, and pass the constant to the right hand.

$$
(\color{brown}{x^2 - 6x}) + (\color{darkorange}{y^2 + 8y }) = \color{darkcyan}{-9}
$$

**Step 2**: Complete the perfect trinomial squares. Add the same values to the right hand.

Warning : Be sure that both coefficients of `$x^2$` and `$y^2$` are equal to `$1$`. Otherwise, first divide each term by this coefficient.

$$
(\color{brown}{x^2 - 6x} + \color{green}{9}) + (\color{darkorange}{y^2 + 8y } + \color{purple}{16}) = \color{darkcyan}{-9} + \color{green}{9} + \color{purple}{16}
$$

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# Example 6 (3/3)

**Step 3**: Factorize the perfect square trinomials, and reduce the constants.

Then $$
(\color{brown}{x^2 - 6x} + \color{green}{9}) + (\color{darkorange}{y^2 + 8y } + \color{purple}{16}) = \color{darkcyan}{-9} + \color{green}{9} + \color{purple}{16}
$$

becomes $$
(\color{brown}{x - }\color{green}{3})^{2} + (\color{darkorange}{y + }\color{purple}{4})^{2} = 16
$$
 **Step 4**: Get the center and the radius from the standar equation. $$
`\begin{gathered}
\text{center: }& C(\color{green}{3},\color{purple}{-4}), \\
\text{radius: }& r=\color{blue}{\sqrt{16}} = \color{blue}{4}
\end{gathered}`
$$

---
# Example 7 (1/2)

**Problem 7**

Find the geometrical elements of the figure with equation given by $$
x^2 + y^2 + 3x + 5y + \frac{9}{4} = 0
$$

**Solution: **
 **Step 1**: $$
(x^2 + 3x) + (y^2 + 5y) = -\frac{9}{4}
$$

**Step 2**: $$
\left( x^2 + 3x \color{red}{+ \frac{9}{4} } \right) + \left(y^{2} + 5y \color{red}{ + \frac{25}{4} }\right) = -\frac{9}{4}\color{red}{ + \frac{9}{4} + \frac{25}{4} }
$$

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# Example 7 (2/2)

**Step 3**: $$
\left( x + \color{green}{ \frac{3}{2} } \right)^2 + \left( y + \color{purple}{ \frac{5}{2} } \right)^2 = \color{blue}{\frac{25}{4} }
$$

**Step 4**: The geometrical elements of the circle are $$
`\begin{gathered}
\text{center: }& C\left(\color{green}{-\frac{3}{2} },\color{purple}{-\frac{5}{2} }\right), \\
\text{radius: }& r=\color{blue}{\sqrt{\frac{25}{4} }} = \color{blue}{ \frac{5}{2} }
\end{gathered}`
$$

---
# Example 8 (1/3)

**Problem 8**

Find the geometrical elements of the figure with equation given by $$
3x^2 + 3y^2 + 4x - 6y - 4 = 0
$$

**Solution**

Note that in this case, both coefficients of `$x^2$` and `$y^2$` are equal to `$\color{darkred}{3}$`. So, the represented figure by the equation _could_ be a circle.

**Step 1**: $$
(3x^2 + 4x) + (3y^2 - 6y) = 4
$$

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# Example 8 (2/3)

**Step 2**: In this case, first we have to divide each term by `$\color{darkred}{3}$` to get coefficients equal to `$1$` for `$x^2$` and `$y^2$`. Then, we get $$
\left( \frac{3x^2}{\color{darkcyan}{3}} + \frac{4x}{\color{darkcyan}{3}} \right) + \left( \frac{3y^2}{\color{darkcyan}{3}} - \frac{6y}{\color{darkcyan}{3}} \right) = \frac{4}{\color{darkcyan}{3}}
$$

this is $$
\left( x^2 + \frac{4}{3} x \right) + \left( y^2 - 2y \right) = \frac{4}{3}
$$
 Next, we complete the squares: $$
\left( x^2 + \frac{4}{3} x \color{red}{+ \frac{4}{9}} \right) + \left( y^2 - 2y \color{red}{ + 1} \right) = \frac{4}{3} \color{red}{ + \frac{4}{9} + 1}
$$

---
# Example 8 (3/3)

**Step 3**: $$
\left( x + \color{green}{\frac{2}{3}} \right)^2 + \left( y - \color{purple}{1} \right)^2 = \color{blue}{\frac{25}{9}}
$$

**Step 4**: The geometrical elements of the circle are $$
`\begin{gathered}
\text{center: }& C\left(\color{green}{-\frac{2}{3} },\color{purple}{ 1 }\right), \\
\text{radius: }& r=\color{blue}{\sqrt{\frac{25}{9} }} = \color{blue}{ \frac{5}{3} }
\end{gathered}`
$$

---
class: inverse, center

PART IV

Applications

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# Example 9 (1/5)

**Problem 9**

A pizza delivery area can be represented by a circle, and extends to the points `$(0,18)$` and `$(-6, 8)$`: these points are on the **diameter** of this circle.
 
 <ol style="font-size:28px">
 <li> At which point is located the pizzeria?
 <li> Find the largest distance of service from the pizzeria.
 <li> Write an equation that represents the circle that encloses the delivery area.
 <li> What is the area of the enclosed surface?
 <li> If your house is at $ (-3,10) $, can you receive a pizza? </li>
 </ol>

---
# Example 9 (2/5)

**Solution**

`$\color{blue}{1.}$` We have to make some remarks in order to solve this problem:

+ The pizzeria is located at the **center** of the circle.
 
 + The center of a circle can be found as the midpoint of a diameter.
 + A diameter of a circle is a segment joining 2 points on the circle that passes through the center.

+ The **midpoint** of a segment with endpoints `$(\color{blue}{x_1, y_1})$` and `$(\color{green}{x_2, y_2})$` is $$
 M\left( \frac{\color{blue}{x_1} + \color{green}{x_2}}{2}, \frac{\color{blue}{y_1} + \color{green}{y_2}}{2} \right)
 $$

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# Example 9 (3/5)

Then, because the endpoints are `$A(\color{blue}{0}, \color{blue}{18})$` and `$B(\color{green}{-6}, \color{green}{8})$`, the pizzeria is located at $$
 C\left( \frac{\color{blue}{0} + \color{green}{(-6)}}{2}, \frac{\color{blue}{18} + \color{green}{8}}{2} \right) = C(\color{green}{-3}, \color{purple}{13})
 $$
 
<hr>

`$\color{blue}{2.}$` The largest distance of service from the pizzeria can be found as the distance to any of the endpoints of the diameter. Here, we are finding the distance between `$A$` and `$C$`: this is the radius of the circle! $$
`\begin{aligned}
\color{blue}{r} & = d(A, C) \\ &= \sqrt{ (0 - (-3))^2 + (18 - 13)^2 } \\
& = \sqrt{9 + 25} \\
& = \sqrt{34}
\end{aligned}`
$$

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# Example 9 (4/5)

`$\color{blue}{3.}$` We have the center `$C(\color{green}{-3}, \color{purple}{13})$` and the radius `$r = \color{blue}{\sqrt{34}}$` of the circle, then the standard equation of this circle is $$
(x - (\color{green}{- 3}) )^2 + (y - \color{purple}{13})^2 = \left( \color{blue}{\sqrt{34}}\right)^2 
$$

this is $$
(x + 3)^2 + (y - 13)^2 = 34
$$

<hr>

`$\color{blue}{4.}$` Because we are working with a circle, the area enclosed can be found as follows: $$
Area = \pi r^2 = \pi \left( \sqrt{34}\right)^2 = \color{red}{34\pi}
$$
---
# Example 9 (5/5)

`$\color{blue}{5.}$` Let's find the distance between my house at `$D(-3, 10)$` and the pizzeria at `$C(-3, 13)$`: $$
`\begin{aligned}
 d(D,C) & = \sqrt{ (-3 - (-3))^2 + (10 - 13)^2 } \\
 & = \sqrt{ 0 + 9 } \\
 & = 3
\end{aligned}`
$$

Because `$9 < 34$`, we know that `$3 = \sqrt{9} < \sqrt{34} = r$`.

In conclusion, as my house is closer to the pizzeria than the radius of the circle, **I can receive a pizza**.