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Welcome to Conic Sections

Samuel Dominic Chukwuemeka (SamDom For Peace) I greet you this day,
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Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Introduction

Conic Sections

Circle

A Circle is defined as the locus of points equidistant (equal distance) from a fixed point.
The fixed point is the center of the circle.
The equal distance is the radius of the circle.

The equation of a circle can be written in any of these two forms:
Standard Form
$(x - h)^2 + (y - k)^2 = r^2$
where:
$x, y$ are the variables
$(h, k)$ are the coordinates of the center of the circle
$r$ is the radius of the circle

General Form
$x^2 + y^2 + cx + dy + e = 0$
where:
$x, y$ are the variables
$c$ is the coefficient of $x$
$d$ is the coefficient of $y$
$c, d, e$ are values/constants

NOTE For:
(1.) $(x - h)^2 + (y - k)^2 = r^2$; Center = $(h, k)$, Radius = $r$

(2.) $(x - h)^2 + (y + k)^2 = r^2$; Center = $(h, -k)$, Radius = $r$

(3.) $(x + h)^2 + (y - k)^2 = r^2$; Center = $(-h, k)$, Radius = $r$

(4.) $(x + h)^2 + (y + k)^2 = r^2$; Center = $(-h, -k)$, Radius = $r$

(5.) $(x - h)^2 + (y - k)^2 = e$; Center = $(h, k)$, Radius = $\sqrt{e}$

(6.) $(x - h)^2 + (y + k)^2 = e$; Center = $(h, -k)$, Radius = $\sqrt{e}$

(7.) $(x + h)^2 + (y - k)^2 = e$; Center = $(-h, k)$, Radius = $\sqrt{e}$

(8.) $(x + h)^2 + (y + k)^2 = e$; Center = $(-h, -k)$, Radius = $\sqrt{e}$

(9.) When converting from Standard Form to General Form, expand. Multiply the binomials and arrange the terms in order.

(10.) When converting from General Form to Standard Form, Completing the Square method is used. Then, arrange the terms in order.

Calculator for Circles

All input values should be integers or decimals only. No fractions
Some of the output values are in fractions and/or radicals. Pick the one you need.
Simplify further as necessary.
  • Given: Standard Form of the Equation of a Circle
    To Find: other details

$(x - $$)^2$ $\:\:+\:\:$ $(y - $$)^2$ $\:\:=\:\:$ $^2$



The center is (, )

The radius is

The General Form is:
$x^2$ $+$ $y^2$ $+$ $x$ $+$ $y$ $+$ $= 0$

  • Given: General Form of the Equation of a Circle
    To Find: other details

$x^2$ $+$ $y^2$ $+$ $x$ $+$ $y$ $+$ $= 0$



The center is (, ) OR (, )

The radius is OR

The Standard Form is:
$(x - $$)^2$ $\:\:+\:\:$ $(y - $$)^2$ $\:\:=\:\:$ $^2$

  • Given: Center, Radius of a Circle
    To Find: other details

The center is (, )

The radius is



The Standard Form is:
$(x - $$)^2$ $\:\:+\:\:$ $(y - $$)^2$ $\:\:=\:\:$ $^2$


The General Form is:
$x^2$ $+$ $y^2$ $+$ $x$ $+$ $y$ $+$ $= 0$

  • Given: Center, Point on the Circumference of a Circle
    To Find: other details

The center is (, )

1st endpoint of diameter is (, )

The radius is OR

The diameter is OR

2nd endpoint of diameter is (, )

The Standard Form is:
$(x - $$)^2$ $\:\:+\:\:$ $(y - $$)^2$ $\:\:=\:\:$ $^2$

The General Form is:
$x^2$ $+$ $y^2$ $+$ $x$ $+$ $y$ $+$ $= 0$

  • Given: End Points of the Diameter of a Circle
    To Find: other details

1st endpoint of diameter is (, )

2nd endpoint of diameter is (, )

The center is (, ) OR (, )

The diameter is OR

The radius is OR

The Standard Form is:
$(x - $$)^2$ $\:\:+\:\:$ $(y - $$)^2$ $\:\:=\:\:$ $^2$

The General Form is:
$x^2$ $+$ $y^2$ $+$ $x$ $+$ $y$ $+$ $= 0$

  • Given: Any Two Points
    To Find: Point on the $y-axis$ Equidistant from the Two Points

1st point is (, )

2nd point is (, )

The point on the $y-axis$ equidistant from the two points is (, )

  • Given: Any Two Points
    To Find: Point on the $x-axis$ Equidistant from the Two Points

1st point is (, )

2nd point is (, )

The point on the $x-axis$ equidistant from the two points is (, )

Formulas for Circles

(1.) Radius, Diameter, Circumference, Area

$ d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] C = \pi d \\[3ex] C = 2\pi r \\[3ex] A = \pi r^2 \\[3ex] A = \dfrac{\pi d^2}{4} \\[5ex] $ (2.) Standard Form of the Equation of a Circle
$(x - h)^2 + (y - k)^2 = r^2$
where:
$x, y$ are the variables
$(h, k)$ are the coordinates of the center of the circle
$r$ is the radius of the circle

(3.) General Form of the Equation of a Circle
$x^2 + y^2 + cx + dy + e = 0$
where:
$x, y$ are the variables
$c$ is the coefficient of $x$
$d$ is the coefficient of $y$
$c, d, e$ are values/constants

(4.) Given: The Center Coordinates of a Circle and an Endpoint on the Circumference of the Circle
The coordinates of the center of the circle = $(h, k)$
The endpoint on the circumference of the circle = $(x_1, y_1)$
The radius of the circle can be found by the Distance Formula
The radius of the circle = $r$
r = $\sqrt{(x_1 - h)^2 + (y_1 - k)^2}$
The diameter of the circle = $d$
The diameter of the circle is twice the radius.
$d = 2 * r$
The second endpoint of the diameter of the circle can also be found
The second endpoint of the diameter of the circle = $(x_2, y_2)$
$ x_2 = x_1 + r \\[3ex] y_2 = y_1 + r \\[3ex] (x_2, y_2) = (x_1 + r, y_1 + r) \\[3ex] $ (5.) Given: The Endpoints on the Circumference of the Circle
$(x_1, y_1)$ = first endpoint of the diameter of a circle
$(x_2, y_2)$ = second endpoint of the diameter of a circle
The center of the circle is found using the Midpoint Formula
$(h, k)$ are the coordinates of the center of the circle

$ h = \dfrac{x_1 + y_1}{2} \\[5ex] k = \dfrac{x_2 + y_2}{2} \\[5ex] $ The radius of the circle can then be found by the Distance Formula using either endpoints
Please refer to Number (3.)

(6.) Given: Any Two Points
To Find a Point on the $y-axis$ Equidistant From the Two Points
Let the first point = $(x_1, y_1)$
and the second point = $(x_2, y_2)$
A point on the $y-axis$ equidistant from $(x_1, y_1)$ and $(x_2, y_2)$ is $(0, y)$
This implies that the distance from $(x_1, y_1)$ to $(0, y)$ should be the same distance from $(0, y)$ to $(x_2, y_2)$
Using the Distance Formula to find the distance from $(x_1, y_1)$ to $(0, y)$ gives:
$ distance1 = \sqrt{(0 - x_1)^2 + (y - y_1)^2} \\[3ex] distance1 = \sqrt{(-x_1)^2 + (y - y_1)^2} \\[3ex] distance1 = \sqrt{x_1^2 + (y - y_1)^2} \\[3ex] $ Using the Distance Formula to find the distance from $(0, y)$ to $(x_2, y_2)$ gives:
$ distance2 = \sqrt{(x_2 - 0)^2 + (y_2 - y)^2} \\[3ex] distance2 = \sqrt{(x_2)^2 + (y_2 - y)^2} \\[3ex] distance2 = \sqrt{x_2^2 + (y_2 - y)^2} \\[3ex] $ The two distances should be the same ...the word, equidistant means equal distance
So, $distance1 = distance2$

$ \sqrt{x_1^2 + (y - y_1)^2} = \sqrt{x_2^2 + (y_2 - y)^2} \\[3ex] Square\:\:both\:\:sides \\[3ex] \left(\sqrt{x_1^2 + (y - y_1)^2}\right)^2 = \left(\sqrt{x_2^2 + (y_2 - y)^2}\right)^2 \\[3ex] x_1^2 + (y - y_1)^2 = x_2^2 + (y_2 - y)^2 \\[3ex] (y - y_1)^2 - (y_2 - y)^2 = x_2^2 - x_1^2 \\[3ex] Expand \\[3ex] (y - y_1)(y - y_1) - [(y_2 - y)(y_2 - y)] = x_2^2 - x_1^2 \\[3ex] y^2 - yy_1 - yy_1 + y_1^2 - (y_2^2 - yy_2 - yy_2 + y^2) = x_2^2 - x_1^2 \\[3ex] y^2 - 2yy_1 + y_1^2 - (y_2^2 - 2yy_2 + y^2) = x_2^2 - x_1^2 \\[3ex] y^2 - 2yy_1 + y_1^2 - y_2^2 + 2yy_2 - y^2 = x_2^2 - x_1^2 \\[3ex] y^2 \:\:cancels\:\:out \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:y \\[3ex] 2yy_2 - 2yy_1 = x_2^2 - x_1^2 - y_1^2 + y_2^2 \\[3ex] y(2y_2 - 2y_1) = x_2^2 + y_2^2 - x_1^2 - y_1^2 \\[3ex] y = \dfrac{x_2^2 + y_2^2 - x_1^2 - y_1^2}{2y_2 - 2y_1} \\[5ex] $ (7.) Given: Any Two Points
To Find a Point on the $x-axis$ Equidistant From the Two Points
Let the first point = $(x_1, y_1)$
and the second point = $(x_2, y_2)$
A point on the $x-axis$ equidistant from $(x_1, y_1)$ and $(x_2, y_2)$ is $(x, 0)$
This implies that the distance from $(x_1, y_1)$ to $(x, 0)$ should be the same distance from $(x, 0)$ to $(x_2, y_2)$
Using the Distance Formula to find the distance from $(x_1, y_1)$ to $(x, 0)$ gives:
$ distance1 = \sqrt{(x - x_1)^2 + (0 - y_1)^2} \\[3ex] distance1 = \sqrt{(x - x_1)^2 + (-y_1)^2} \\[3ex] distance1 = \sqrt{(x - x_1)^2 + y_1^2} \\[3ex] $ Using the Distance Formula to find the distance from $(x, 0)$ to $(x_2, y_2)$ gives:
$ distance2 = \sqrt{(x_2 - x)^2 + (y_2 - 0)^2} \\[3ex] distance2 = \sqrt{(x_2 - x)^2 + (y_2)^2} \\[3ex] distance2 = \sqrt{(x_2 - x)^2 + y_2^2} \\[3ex] $ The two distances should be the same ...the word, equidistant means equal distance
So, $distance1 = distance2$

$ \sqrt{(x - x_1)^2 + y_1^2} = \sqrt{(x_2 - x)^2 + y_2^2} \\[3ex] Square\:\:both\:\:sides \\[3ex] \left(\sqrt{(x - x_1)^2 + y_1^2}\right)^2 = \left(\sqrt{(x_2 - x)^2 + y_2^2}\right)^2 \\[3ex] (x - x_1)^2 + y_1^2 = (x_2 - x)^2 + y_2^2 \\[3ex] (x - x_1)^2 - (x_2 - x)^2 = y_2^2 - y_1^2 \\[3ex] Expand \\[3ex] (x - x_1)(x - x_1) - [(x_2 - x)(x_2 - x)] = y_2^2 - y_1^2 \\[3ex] x^2 - xx_1 - xx_1 + x_1^2 - (x_2^2 - xx_2 - xx_2 + x^2) = y_2^2 - y_1^2 \\[3ex] x^2 - 2xx_1 + x_1^2 - (x_2^2 - 2xx_2 + x^2) = y_2^2 - y_1^2 \\[3ex] x^2 - 2xx_1 + x_1^2 - x_2^2 + 2xx_2 - x^2 = y_2^2 - y_1^2 \\[3ex] x^2 \:\:cancels\:\:out \\[3ex] Collect\:\:like\:\:terms\:\:in\:\:x \\[3ex] 2xx_2 - 2xx_1 = y_2^2 - y_1^2 - x_1^2 + x_2^2 \\[3ex] x(2x_2 - 2x_1) = x_2^2 + y_2^2 - x_1^2 - y_1^2 \\[3ex] x = \dfrac{x_2^2 + y_2^2 - x_1^2 - y_1^2}{2x_2 - 2x_1} \\[5ex] $

Formulas for Conic Sections

Parabolas

(1.) Standard Form of the Equation of a Circle

Gay Lussac's Law of Combining Volumes

$\rightarrow$ Attributed to Joseph Gay-Lussac
$\rightarrow$ Describes the combining volumes of reacting gases (gaseous reactants) and gaseous products
$\rightarrow$ States that at a constant temperature and pressure, the volumes of gaseous reactants and gaseous products occur in simple whole number ratios.

References

Chukwuemeka, S.D (2019, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.chukwuemekasamuel.com

Bittinger, M. L., Beecher, J. A., Ellenbogen, D. J., & Penna, J. A. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ ed.). Boston: Pearson.

Coburn, J., & Coffelt, J. (2014). College Algebra Essentials ($3^{rd}$ ed.). New York: McGraw-Hill

Sullivan, M., & Sullivan, M. (2017). Algebra & Trigonometry ($7^{th}$ ed.). Boston: Pearson.

CrackACT. (n.d.). Retrieved from http://www.crackact.com/act-downloads/