[Math 20] Lec 1.5 Lines and Circles
Summary
TLDRThe video provides an in-depth exploration of the equations of lines and circles. It covers essential concepts like the slope of a line, point-slope form, slope-intercept form, and the general equation of a line. The video also discusses perpendicular and parallel lines, finding equations of lines passing through specific points, and the properties of tangent lines to circles. The explanation includes formulas, step-by-step examples, and a focus on how to find the slope and intercept of lines, as well as the equation of a circle and its tangent.
Takeaways
- 📝 The video explains the equations of lines and circles, starting with a review of a unique line and its slope.
- 📏 The slope of a line is determined by the formula: slope (m) = (y2 - y1) / (x2 - x1), representing the steepness of the line.
- 🔄 To find the equation of a line, use the point-slope form: y - y1 = m(x - x1), which is crucial for describing lines.
- 🔗 The slope-intercept form of a line is y = mx + b, where b is the y-intercept of the line.
- 🔍 Parallel lines have the same slope, while perpendicular lines have slopes that are negative reciprocals of each other.
- 🔄 The video demonstrates how to find the equation of a line that passes through two points or is perpendicular to another line.
- ⚪ The equation of a circle is given as (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.
- 📏 The distance formula between two points is used to relate points on a circle to its center.
- 🧮 The general form of a circle’s equation can be expanded to find specific values for the center and radius.
- 📐 Tangent lines to a circle intersect it at exactly one point and are perpendicular to the radius at the point of tangency.
Q & A
What is the slope of a line and how is it determined?
-The slope of a line represents its steepness and is determined using the formula: slope (m) = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two points on the line.
How do you find the equation of a line using a point and the slope?
-The equation of a line can be found using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
What is the slope-intercept form of a line equation?
-The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept, the point where the line crosses the y-axis.
How do you find the equation of a line that passes through two points?
-To find the equation of a line passing through two points, first calculate the slope using m = (y2 - y1) / (x2 - x1), then use the point-slope form y - y1 = m(x - x1) with one of the points.
What does it mean for two lines to be parallel?
-Two lines are parallel if they have the same slope, meaning the slopes of the two lines are equal (m1 = m2).
What is the relationship between the slopes of two perpendicular lines?
-Two lines are perpendicular if the product of their slopes is -1. This means if one line has a slope of m, the other line has a slope of -1/m.
How can you find the equation of a line perpendicular to another line and passing through a given point?
-First, find the slope of the given line and take the negative reciprocal to get the slope of the perpendicular line. Then, use the point-slope form with the given point to find the equation.
What is the definition of a circle in a plane?
-A circle is defined as a set of points in a plane that are equidistant from a fixed point called the center. The distance from the center to any point on the circle is the radius.
How do you write the equation of a circle given its center and radius?
-The equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r².
What is a tangent line to a circle, and how is it characterized?
-A tangent line to a circle is a line that touches the circle at exactly one point. At the point of tangency, the line is perpendicular to the radius drawn to the point of tangency.
Outlines
📐 Introduction to Equations of Lines and Circles
The video begins by introducing the concept of equations related to lines and circles. It recalls the importance of the slope in determining the steepness of a line. The formula for calculating the slope using two points on a line (x1, y1) and (x2, y2) is provided, along with a detailed explanation of how to derive the equation of a line using the point-slope form.
✏️ Deriving the Equation of a Line
This section explains how to find the equation of a line that passes through two points. The video demonstrates using the slope formula, substituting into the point-slope form, and finally converting it into the slope-intercept form. An example is given, showing how to derive the equation of a line using specific points and the negative reciprocal of the slope.
🛤️ Perpendicular Lines and Equations
The discussion shifts to lines that are perpendicular to each other, explaining that two lines are perpendicular if the product of their slopes is -1. The process to find the equation of a line that is perpendicular to another given line is outlined, using a specific example to illustrate the method.
⭕ Introduction to Circles and Their Equations
This section defines what a circle is and introduces the equation of a circle in the plane. The center-radius form of the circle's equation is presented, along with a brief explanation of how to derive it using the distance formula between two points. The general form of the circle's equation is also introduced, providing a method to determine the center and radius.
📏 Finding the Equation of a Circle
Here, the video walks through the process of finding the equation of a circle given its center and a point on the circle. It involves substituting the given center and point into the center-radius form and solving for the radius. The general equation of the circle is then provided, showing how to relate it to the center-radius form.
🔗 Tangent Lines to Circles
The final section explores tangent lines to circles, which intersect the circle at exactly one point. The video explains how to find the equation of a tangent line to a given circle at a specific point, using the perpendicularity property of the tangent and the radius. An example is provided to demonstrate this method.
📚 Conclusion on Lines and Circles
The video concludes with a summary of the key concepts discussed, including equations of lines, circles, and tangent lines. The instructor hopes that the viewers have understood the material, and thanks them for watching.
Mindmap
Keywords
💡Slope
💡Point-Slope Form
💡Slope-Intercept Form
💡Parallel Lines
💡Perpendicular Lines
💡Circle
💡Center of a Circle
💡Radius
💡Tangent Line
💡General Form of a Circle
Highlights
Introduction to equations of lines and circles, emphasizing the importance of understanding the slope and determining the uniqueness of a line.
Explanation of slope as the steepness of a line, and how to calculate it using the formula (y2 - y1) / (x2 - x1).
Introduction of the point-slope form equation: y - y₁ = m(x - x₁), highlighting its importance in describing a line given a point and its slope.
Derivation of the slope-intercept form equation: y = mx + b, and its significance in finding the equation of a line with a known y-intercept.
Discussion of how to find the equation of a line that passes through two points using the slope-intercept form.
Explanation of parallel and perpendicular lines, including the conditions for parallel lines (same slope) and perpendicular lines (product of slopes equals -1).
Detailed steps to find the equation of a line perpendicular to a given line and passing through a specific point using the negative reciprocal of the slope.
Introduction to circles as a set of points equidistant from a center point, and the equation of a circle: (x - h)² + (y - k)² = r².
Application of the distance formula to derive the equation of a circle with a known center and radius.
Explanation of the general form of a circle's equation and how to convert it to the standard form to find the center and radius.
Worked example on finding the equation of a circle given a center and a point on the circle, demonstrating the use of the standard form.
Introduction to tangent lines to circles, defining a tangent line as one that intersects a circle at exactly one point.
Illustration of the special property of tangent lines: the radius at the point of tangency is perpendicular to the tangent line.
Example problem on finding the equation of a tangent line to a circle at a specific point, including deriving the slope and using the slope-intercept form.
Conclusion summarizing the key concepts of lines, circles, and their interactions, reinforcing the importance of the equations and methods discussed.
Transcripts
hello so in this video we're going to
talk about the equations of
lines and circles
so let us recall come on
unique line so determine
is a unique length
on the line and the slope of the line
so pakistan's slope of the line that is
the steepness
of airline so back at nothing
steepness of a line so let's say this is
your point
when we say steepness or the slope
uh or because say some point maraming
lines
um so parama specify nothing
we have to determine the slope of the
line
and to get the slope of a line so let us
recall so let's say this is your line
you have two points
x to y two and x one y y1
you can compute for the slope m using
the formula y2 minus y1 all over
x2 minus x1 or the change in y divided
by the change in
x and how do we get the equation of a
line so he's having an attitude
and at a point in the slope so let's say
the slope of her line is
m and x not y
naught is a point on the line so pan
another major describe it online at all
so let's say we have a generic point on
your line so
how are these x and y related
so let me tell you a formula in a slope
so sabina
and this that the slope is equal to this
one
so let's say um
this is your x2 y2
so this slope is equal to y minus
y naught divided by x minus x naught
you get this one and if you multiply
both sides by this expression
x minus x naught we get this
y minus y not equal to m times
x minus x naught equation
and we call that equation the
point-slope form the initial action a
point-slope form because meanwhile
point x not y-nut and the slope m
so take note of this this is very
important
and your next a very important number
discuss not then
slope intercept of form so melancholine
and intercept in particular y-intercept
number line
so let's say um this point is on the
line l so
homo observing the x coordinate is zero
meaning
b is the y in percent and using the
point slope form which is y minus y
naught equal to m
times x minus x naught
this is her x not y naught but bin laden
0 so x naught we will get this
expression or this equation and y
equal to f x plus b negative d
naught in a slope intercept form
this is a very important equation
from this equation
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the slope intercept form and the general
form so general form
so give it an opinion to find the
equation of
a line so let's say find the equation of
a line that passes through these
two points so he landed at a point and
we need
a slope and
and slope is y two minus y one
over x two minus x one e number one
point
slope you will get the equation of a
line so let's say this is
x one y one this is x to y two
so we get two minus
four divided by five minus
negative three so the slope is negative
two over
y naught equal to m times x minus x
naught
so i don't know equation
so for x not y naught it's just pick one
point
kites in a giant guy cp guide cq
uh pili in a language so this will serve
our x naught
and this is our y naught so x naught is
negative three
y naught is a four and the slope is
negative one fourth
so the equation is y minus
four equal to negative one fourth
x minus
negative 3
distributed
y equal to negative 1 for it
uh this is another way of doing it
direct and a y equal to m x plus b
slope negative one four so the equation
is y
equal to m x plus b or
y equal to negative one fourth x plus b
so allah means b to get the equation of
the length
so the doing one you just pick one point
either this one or this one
or for anonymous
so this particular point must satisfy
this equation because this point is on
the line
so at x y equal to five to one
a bus you get two equal to negative one
fourth x plus b
or two is negative five four x plus b
so b is two plus five fourths
or thirteen fourths so equation online
mo
is uh plug in attendee to c b
y equal to negative one fourth x plus 13
over four so we get the same
answer so you know two ways of doing
at least two ways of doing it just
in perpendicular lines so very important
to y equal to mx
plus because slope when you solve for y
the coefficient of x
is the slope of the line so we say that
two lines are parallel if
they have the same slope and if it's m
of
l one equal to m of l two the slope of
line one is this the slope of line two
is this
and two lines are perpendicular
if the products of their slope is
negative one
or whether it's
l1 is perpendicular to l2 okay or l1 and
l2 are perpendicular
l1 and l2 are parallel
so now of course an example
so find the slope intercept of form and
each one in
slope intercept form that is y equal to
m
x plus b of the equation of the line
that passes through one one and
perpendicular to l
so you know line nut and passes through
p
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um y
y equal to two x plus
okay so you're missing line if i
yes using your slope num l
so the slope of l
observation y equal to m x plus b
so the slope of l is 2
and you want this missing line to be
perpendicular to l
so the slope of this missing line is
negative reciprocal of the slope of l
which is negative one half
so in equation and online is y equal to
negative one half x
plus b
equation this point
passes through this missing line you
know desired line at it
so meaning this point must satisfy this
equation
if you plug this particular x and y
in here the operating equation
so at x y
equal to 1 1 and the best we get one
equal to negative one half times one
plus
b or b is one plus one half
where b is three over so the equation of
the line
is y equal to negative one half x
theta and b is three over two
circles so let us define what a circle
is or
at least so a circle
is a set of points in the plane
equidistant from a fixed point
that fixed point is called the center so
in this case ethereum center net
blue red
highlights
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circle at imbabat point
little circle is equidistant to the
center
distance
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radius
so if the center is h k and the radius
is r
the equation is x minus h squared
plus y minus k squared equal to
r square so pana na ginga non explain
langnaden briefly
so let us recall the formula for the
distance between two points
so young distance between x1 y1
and x2 y2
is uh let's say d equal to
x2 minus x1 squared plus y2
minus y1 square okay
so let's say your circle now then
ayan and the center is h k
and the reduce is r so you want to
relate manga points
circle so let's say you have an
arbitrary point on the circle
x y so sabinate
circle and distance from the center
is r so we're going to use this distance
formula
so r is equal to the square root
of x minus
h squared
plus y minus k square
and voila when you square both sides of
this equation
when you expand this one tapas nila gay
musa left la ha
zero habilong said in general form
is something like x square plus y
squared plus
uh a constant times x
plus a constant times y plus a constant
equal to
zero so union general form or a general
equation and circle
all although in the obvious
center
regious form
because from the equation or from the
form itself you can find the results
in the center so from here
pano nyo malama new
or you can use the formula
uh h is equal to negative d
over 2 e is
negative sorry
k is
negative e over two so i need to divide
mulan atom by negative two to get
h at the demand divide by negative two
to get
k atom regions
so r square is equal to h square
plus k squared minus f so that is r
squared so go what i you know
examples so find
an equation for the circle with center
nato and passing through this point so i
know that your center religious form
so it's x minus h square plus y
minus k squared equal to r square
so our h k in this case is
two negative three so papa literally
nothing see h none two and k by negative
three
so x minus two squared plus
y minus negative magnitude plus
so say r squared
is a problem the circle passes through
negative one one so this negative one
one must satisfy this equation
so we'll plug in another nine
so we just replace x by negative one and
y by one
r square so r squared is equal to
negative three square kasama is negative
say any square so you get nine
using the man four square is sixteen
so r square is twenty five seventy one
so the equation of the circle is
equal to 25
so your new equation
so tangent lines to circles so
a line that intersects a circle at
exactly
one point is called a tangent line to
the circle
so example this one this line is tangent
to the circle because
it touches the circle at exactly one
point
and uh measuring special properties
the point of tangency is perpendicular
to your
tangent line okay
so as i illustrated in this figure
so perpendicular
perpendicular segment
so let us determine the slope intercept
form
of the equation of the line that is
tangent to the circle with this
equation at zero two
um provide time and sketch all the hindi
to accurate sketch huh
rough sketch lung so let's say that is
our
circle and this is
assambus is zero two let's say this is
zero two
circle so x square
plus y square
okay so h naught then
divide this by negative two you get one
our k is divide this by negative two you
get negative two
so r squared is h squared plus k
square minus f or
h squared is one plus four and f not
then is
negative twelve
so this is
so let's say
foreign
drawing
slope intercept form of the tangent line
so equation of tangent line
tl for tangent line is y equal to m x
plus b
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is
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is a line so b
is two so b is equal to b
so y equal to m x plus two
slope nalan ankula so this
gamma and tangent line so sabinate
if you connect the lines the
the points of tangency and the center of
the circle
so let's say it in blue line so
perpendicular
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for the slope of the blue line
okay so anion formula y minus y
2 minus negative 2 divided by
x minus x
or a negative four so
in the hand of nothing slope no tangent
line is a negative reciprocal of the
slope of the blue line
but the negative reciprocal of this is
one-fourth so you know nothing equation
of tangent line is
y equal to one-fourth x plus
plus two
and we can actually check this
and circle it type not then so it's x
squared
plus y squared minus
2x plus 4y
minus 12 equal to zero
so it must be x square plus y squared
up in a y equal to
[Music]
one-fourth x
one-fourth x plus two
yes as you can see ion sharp tangent
shall high
circle at point zero two so c a one zero
two
and that ends our discussions on lines
and circles so serena gets nino
goodbye and thank you
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