Intro to Circles - Conic Sections, Geometry, Mathematics Video Lecture | Crash Course for JEE (English)

Now
that you know what the conic
sections are and why they're
called conic sections, let's
see if we can understand the
actual equations of the conic
sections a little bit better,
and use that knowledge to be
able to at least recognize them
when we see the equation, and
then if we see the equation,
know how to plot them.
So the first one we'll start
off with is the circle.
And you've probably seen this
for many years already,
but I'll review, if you
don't know it already.
The general form of an
equation, or the general
equation for a circle, is x
squared plus y squared is equal
to r squared, where r is
the radius of the circle.
And this would be a circle
centered at the point 0,0.
So I'll just draw this kind of
a general graph right here.
So that's the x-axis, that's
the y-axis, the circle will
look like this-- let me do it
in a different color-- the
circle would look something
like-- make sure it's a
circle-- well, it's supposed to
be centered at 0,0, but that's
close enough-- so its center
will be right here, and the
radius if you go from the
center to any point along that
circle will have a distance of
r, so if you go from there to
there it's r, from there to
there it's r, from there to
there it's r, and to some
degree, this formula, all it is
is an extension of the distance
formula, which is really just
the extension of the
Pythagorean Theorem.
So for example, the distance
formula, if I want to know the
distance between some point x,y
and the point 0,0, what you do
is, you take the difference of
the x's-- so x minus 0-- you
square that, and then you add
that to the distance between
the y's squared-- so that's one
y point minus 0y-- y is equal
to 0-- square that, and that is
equal to the distance squared.
So if you simplify this, x
minus 0 squared, that's just
x squared plus-- and this
is just y squared is equal
to distance squared.
So essentially, this equation
is the plot of all points that
are exactly d away, a distance
of d away, from the point 0,0,
and that's just a circle.
And I'll let you think about
it, I think I actually showed
this to you in a distance
formula video, but the distance
formula just comes out of
the Pythagorean Theorem.
And if that's not completely
obvious for you, just think
about a little bit, and
it'll hopefully become a
little bit more obvious.
But anyway, this was probably--
you probably already knew this,
and actually just to hit the
point home, if we had the
equation like this x squared
plus y squared is equal to 9,
the graph of this circle
would look like this.
So that's the x-axis, that's
the y-axis, then draw the
circle itself, circle looks
like-- close enough, and then
the distance or the radius from
the center of the circle,
that's going to be 3.
There's a 9 here, why
isn't the radius 9?
Oh, that's because this
is the radius squared.
Remember the original formula I
just showed you. x squared plus
y squared is equal
to r squared.
So this right here is r
squared, so if r squared
is equal to 9, than
r is equal to 3.
It can't be minus 3.
I mean, it could, you can't
have a negative radius, or if
you did you're just going in
the other direction, but
it's the same thing.
So the radius is equal to 3.
So that's a circle, and that's
pretty straightforward, but in
a lot of algebra classes, they
complicate the issue a little
bit by shifting the circle.
So let's just shift
this circle.
Instead of having-- so
let me just rewrite it.
So the unshifted circle was x
squared plus y squared is equal
to-- let me write it this way--
is equal to 3 squared, that's
the same thing as 9, and let's
say that the new circle, the
shifted circle, is x minus 1
squared plus y plus 2 squared
is equal to 3 squared.
Now all of a sudden this looks
really complicated and daunting
and all the rest, but all you
have to recognize is, we just
substituted an x minus 1 for
the-- whoops, messed
up my pointer.
We just substituted an x
minus 1 for the x, and
we just substituted a
y plus 2 for the y.
So it has the same basic
pattern as of this circle, and
the fact that we added or
subtracted numbers from the x's
and the y's tells us that we
shifted the circle, and so the
next obvious question is,
where did you shift it to?
And your impulse might be, oh,
well maybe I shifted it to,
instead of the center being
0,0, your intuition might be
to say that, well, now the
center is at negative 1,2.
And you'd be almost right,
except you'd be exactly the
opposite of the correct answer.
The new center is now x is
equal to positive 1 and
y is equal to minus 2.
And that might be unintuitive
for you at first-- and you
might want to watch some of the
videos, I think I've done them
already, or I've always
intended to, on shifting
functions-- but the way to
think about it is the center
here is x is equal to 0.
So when x and y is equal to 0,
x squared plus y squared is 0,
you're exactly 0 away from the
center, or we're at the center.
So now if we want to be-- if we
want x to be 0 away from our
new center, this term
has to be equal to 0.
And if-- just like when x was
equal to 0, this term equaled
0, so now for us to be at the
center of our new circle, so to
speak, this term has to be 0.
So the new center has to
be at x is equal to 1.
Likewise, this has to be
0, and so the center
is at y is equal to 2.
Another way to think about it
is this-- let's say when y,
you know what happens here
when y is equal to 2.
Whatever part of the circle
we're in when y is equal to 2.
There's some part of the circle
we're at, in fact I could draw
it, when y is equal to 2.
Let's say this radius is 3 when
y is equal to 2, we're probably
right around there
on the circle.
We could be there or
we could be there.
Now we're shifting the circle,
so now instead of being at 0,0,
we're going to be at 1 minus 2.
So now we're going to be at the
new center is x is equal to 1,
y is equal to minus 2, the new
center is there, and if I were
to draw the new circle, it
would look something like this.
I'm going to try my best
to draw it still as a
circle and show you
that it's been shifted.
No, that's not good.
Let me draw it like-- I
pressed the wrong button.
That's not good.
Let me draw it right there.
That's close enough.
I don't have to keep doing it.
So what we did is we shifted
this circle down 2,
and to the right 1.
So if we take its center
point, we went down 2
and to the right 1.
And so if you think about up
here, when y was equal to 2, we
could have been at this point
or this point, the kind of
equivalent points of the new
circle are going to be here.
Are going to be roughly here,
where you're shifting it
down and to the right.
And in order to have that
same behavior in the circle
there, this whole thing
should be equal to 2.
So that same point on the
circle, if this whole thing is
going to be equal to 2--
because it's going to be the
same kind of behavior in the
equation, and I hope I'm not
confusing you there-- then the
new y has to be 0, and
you see it there.
At both of these points
now, y is equal to 0.
So I know that's a little
unintuitive, but I want you to
sit and think about that a lot.
I mean, you could just memorize
it, that it's the opposite,
when you have x minus 1 and y
plus 2 that it's actually you
shifted to x is equal to-- the
center is now 1, minus 2, or
you could memorize, if you
like, that what makes this 0
and what makes this 0, and
that's your new center.
But I really want you to
think about it, this
is really a shift.
And of course, if you were
to graph it, you were to
get this thing up there.
Anyway, let me see how
much time I have.
Actually I didn't even
keep track of the time.
I'll leave you there, I'll
continue this in the next video
where I'll talk a little
bit about ellipses.