One and only circle passes through any three points drawn in the plane (Theorem and proof) Video Lecture - Class 9

close for welcome welcome to this
presentation from rising pearl hum apke
host after destroy those though today be
a discussing series 10 jaha babe we are
talking about circles eyes have babies
at number 11 each case presentation
macabre hug botzwogga-beesh
now those Kamala man that from The
Wizard 1 through 10 up that we have seen
a lot of the the fundamentals of circle
properties of circle and also those that
we have proved lot of important theorems
so the Hamas rocket they are following
us along our ghost though we are going
to prove a new theorem that says there
is one and only one circle one and only
one circle passing through three given
non collinear points
now those the webisode 10 may come
nebith Keith he circles and points key
relationship cavalry man I've got
America's eight-point BJT hey how many
the hit he last webisode main key hum
infinite circles draw cursor plane that
passes through one given point I got him
a dew point desert the heroes though
still HMDA the hit held a key key key we
can draw in finite circles that passes
through any two given points so also our
chi is a theorem this theorem is telling
us key there is one and only one circle
that can pass through three given non
collinear points so he had also ASCII
theorem and we are going to find out
casts occurred to him as a proof cursor
player such a lathe also let us get
started so jess I am always curtain
heparins let us understand what the
theorem is telling us so there is one
and only one unique circle passing
through three given non collinear points
so you have to go snuggle Marcos has
take a k-qui between random non
collinear points agar plate a ho so in
this case also we have taken these three
points at will be shaped wavy distance
key you can take three given points Jodi
Hamel acne had also that the three given
points the three given points they must
be actually non collinear
of Kamala may collinear come at la Vega
repair as a client rocket a ho who's
lying
OOP
in point little the Equality number of
points little so I got a second camera
Christine point here or and all of these
three points other would they lie on the
same line then ham in pinpoint go home
Callahan basically collinear because all
of these three points they lie on the
same line so humming habit hammer at
MiraCosta ham 18 non collinear points to
here so now the cigar opposite in Kobe
non collinear points late a ho
so theorem is telling us that we must be
able to draw one and only one circle
that passes through each of these three
points some are 18 points goes through
this this and this so the theorem is
telling us keep this is the only circle
there is one unique circle this that can
be drawn that passes through each of
these three points or SK lava there can
be no other circle that can be drawn so
this is what the theorem is telling us
that this is the only unique circle that
can be drawn passing through our three
non collinear points to hominid hemara
is example may we have taken so this is
now we are going to find out how we can
prove this but escape the hell are those
though we want to just talk about one
very important thing about circles that
you already know by now but we still
wanted to discuss this tacky we
reinforce that and was cabal hum a move
carrying a key how to prove this
particular theorem so you have a goes to
what we have done is that man lee jaehun
18 circles rocky here so chile conquered
let us call this as o 1 and this as o to
the center of these so you have a go
stop notice karaage paheli what we have
done
here's a circle had Rostow so hum nay we
have taken a centre o 1 and we have
taken a radius R silly say R 1 quatre so
occur those Co a hamari centered her o
one and if we take the radius
this key lengths are one so we get this
circle silicic at a circle one this is a
circle one then those four what we have
done is hum the same center or one Li
here and man Lydia this time we draw a
radius like this or is a cat there are
two usual radius of the scope and it's
circle cook at the head second circle
right
so those though what we want to tell you
is this agar up a unique centre lathe a
home or up different radiuses draw at
the hook so I put different different
circles m'lady right so our op o one
center link a graph our one radius draw
at a ho I go this unique circle Milegi
Circle one agar up o one Center Lake a
unique radius are too late a ho then you
will get a circle to now those are you
happy what we have done hum nay extent
early have o to and you happy those two
what we have done is hummed a same
radius Li has our one so those two hum
decay and this is our third circle so
mad lobby hadass though Kyogre hum same
Center later here different radiuses
later a man allegoric circles milky
he'll also those who hey be truhair yoga
ham different centers later or same
radius lee there will be homologue alex
circles will be here so important body
hey Rosco
that a girl if we have if we have one
Center if we have one Center plus one
radius
I got America's egg center here and if
we have one radius we will get one
unique circle so yeah those are
something this is something that I've
quality Mahlum here but we still wanted
to mention this because they
logic is going to be used actually asked
is theorem key provement key same Center
and different radius Lena say r1 r2 Lane
is the same Center different radius of
Cola Gallic circles Milegi different
centers same radius Lane is still a
koala Gallic circles me lately
Muttley we had respect for one centre
and one radius there is only one unique
circle that is available so those are
age a particular cheese's a hum bock
saga is his philosophy Co is logical use
cutting edge keep provement so jelly
those are now with all of that said let
us get into the proof itself so Hamada
pass team points the hlaa conquer the
hell it escaped them names so manly ji
hamari pass a points a B and C now it's
proof me though so there is some
construction which is involved that you
pay attention to this construction and
then we will get into the proof the
param character hello so this is what we
do first
sokka hella ham a line each of my 18
points a a a B and C we draw two lines
like a b and b c hum a Bicol a second
cage or the hen and hum bc co a second
cage or there so our Kamala Nagar
octopus team points our teen line
rockers at the hole so Tina Mikasa we
want to draw two lights so how many
happy we have drawn a B and we have
drawn BC is construction Chabad those
two what we want to do is that a B hum
we want to draw a perpendicular bisector
of a B so yeah Hammadi
line hey a B is key we want to draw a
perpendicular bisector mutlar key we
want to draw a line Chili's is line cook
at the hairline
L so line L is such a line which is
perpendicular bisector of a be a B right
Mac lucky this is equal to this so a
beak is your midpoint
one said we draw a 90-degree line like
this and he said at the hairline L
similarly goes the homology do three
side here BC hum escape he perpendicular
bisector Drucker Natura there
so now Chile is a cat here this line
cook at the hemline M and he had a
similar Li those thought this is equal
to this and this angle is 90 degrees now
those though akumal um have from Japan
parallel aligns angles cavalry me
discuss corrected agar aapke paas will
be dough line a hair like this line and
maybe at this line other year don't know
line parallel in here then is key
perpendicular and and perpendicular from
this line they will intersect right
similarly the Swagelok a pass though
parallel lines are like this is key
perpendicular and is key perpendicular
they'll be always parallel so you have
because the clearly a B and BC they
intersect at point B is kamala rior
those a perpendicular bisector of a be
line L and line M which is perpendicular
bisector of BC yet ona they will
intersect at some point right the
conquer the coaster is point Co so this
is this is this point you have a line L
and line M they intersect right so a VM
Calcutta hello Oscar Chile is point Co
Vietnam they then let us call it Oh a VM
Calcutta he goes to rejoin Oh a right
and now we also want to join OB and OC a
second Oscar hum may join key here Oh a
OB and OC now those though what we can
do is that we can actually draw a circle
that passes through a B and C so those
are push okay T why is this true why
it possible to draw a circle like this
right but jelly I hear now we are going
to actually discuss the proof itself so
you happy those two what we have is that
hamari JT point a yeti point B yeti
Point C correct and he had was the line
L joking perpendicular bisector of a B
right
he had 90 degrees similarly line he line
m and this is a perpendicular bisector
of BC and hum Monta had also P L and M
they intersect at o husqvarna be joined
Oh a OB and OC so you have a goes to the
reason we can draw one and only one
unique circle is this
so how me madam Hickey has a point oh
hey this is a unique point right Q
Kyogre octopus will be though lino here
if you have any two lines a Cline like
this and acquire line like this these
two lines will intersect only at one
unique point so line L and line m they
intersect at point O so this point O is
a unique point
now those cottage is notice TJ aptech
okay is a point O here it lies on line L
right like in line L key a line L
actually have perpendicular bisector of
a B so it's kamek la vieja da sake o a
and OB are equal so o a is equal to OB
so after okay why because was the point
oh it lies on perpendicular bisector of
a B and her my Mohammed also we know
this property he any point on the
perpendicular bisector is equidistant
from both these two points six comma
Clavius Oscar he other manager capaci
line hey like this right Alicia got her
XY or hum is cubic perpendicular
bisector drop
something like this right
so now aghanim is lying pxy lanky
perpendicular bisector rockers
potentially sick at the hairline p then
those though is lying ke upar agar aapko
AV point little manly G at this point
yeah at this point you're at this point
each of these points they are equal
equidistant from x and y McClatchy agree
a point I managed at P 1 then P 1 X is
equal to p1 y occur a point happy 3
so then goes though p3 X is equal to p3
why we know this though so this is a
theorem that we are aware how a madam
has theorem guitar remain so now so
because the point O here it lies on line
L jokey perpendicular bisector hair of a
be system at level also Oh a will be
equal to OB so yeah hadassah hamari
paheli condition now of the hovitos
though also jump point
oh hey that lies on line em but line NK
here goes the line m is nothing but a
perpendicular bisector of b.c.e schemata
via those other o lies on line m so OB
must be equal to OC
so OB must be equal to OC shall he take
a take a condition number 2 so those
hamari a grab a window no condition
Camilla say ho 1 & 2 so hum day ahead
ki jo o a hair is equal to OB and your
OB has equal to OC so kanaka maole ba da
Scotty a jockey no length hair o a OB
and OC
these are all equal lengths say Concord
and Australia let us assume k jo be a
lengthy measure her cello secretary our
unit system at Lafayette Oscar with O as
Center and these lengths are as radius
we can draw a circle again a massacre
click circled rocker there then that
circle because o point unique point it
is intersection of L and M so this is a
unique point and hamari pass we are we
are taking a unique point and we are
taking a fixed radius R so hum if you
draw a circle we can draw one and only
one circle we can draw one and only one
circle one and only one circle we can
draw that actually passes through a B
and C so those so this is the proof how
many should watch it here with three non
collinear points opt non-collinear
points or hamnida sake we can get X
enter and we can get a unique radius and
a unique center and neck unique radius
Lake a we can draw one and only one
unique circle
joki only no point say it goes through
all the points were it that that circle
it did actually it that go circle goes
out to here all of those three points
came through self so those are now let
us actually go ahead and write it
appropriately exactly AB you will know
is a classical cap go step by step look
near so hum should watt cut the hair by
what is given we have three non
collinear points a B and C as shown a B
and C whose kabab to prove the make a
proof conahey we need to prove that
there is one and only one circle that
passes through three points a B and C
melibea Hecky a and only one circle is
there that passes through a B and C then
those two comes the very important
construction discovered mohammed khalid
so we joined a b and b c then we draw
perpendicular bisector of a
right say lined L just say I'm like AHA
dosto so now we draw perpendicular
bisector of BCC line M since a B and BC
are not parallel they're perpendicular
line L and M will not be parallel so
line L and M will intersect and let us
say the intersect at point O correct so
it's come at la vía who are those though
now we can say that we know if we take
any point on the perpendicular bisector
of a be that point is equidistant from
both a and B so oh tujhe point hammer is
on line L is come at la vieja those
though he other o is on line L so we
have array equals to OB right so it's
come at la vieja so Joe Oh a here this
length is actually equal to this length
UK Joe point
oh here it lies on perpendicular
bisector this is a perpendicular
bisector okay we say cat head was the
condition one right then similarly if we
take any point on perpendicular bisector
of BC that point is equidistant from
both B and C and how mamalu my o is on
line M also so he has a VC that OB is
equal to OC he say ham cat the heck
condition - right so hamari Pazzo's da
da da no conditions
so from this condition some ability here
that oh a is equal to OB is equal to OC
right now since o is such a point that
is equidistant from all three points a B
and C so we can draw a circle with
Center oh oh Co Center lay the way we
can draw a circle with radius Oh a that
will pass through B and C because Oh a
OB and OC is a V equal Sagara moco
centered Lake a and Omega radius look at
the Rec circle draw car there
so that circle will pass through B and C
because a OB and OC they are of
well lengths right so finally duster I'm
Canada the hankie Oh is a unique point
since oh is a unique point because point
of intersection of perpendicular
bisectors a B and a C and we saw that Oh
a equal to OB equal to AC we can say hum
Kazik when there is one and only one
circle that passes through all three non
collinear points
so those the ham naked aha last
presentation Mathematica Kyogre up eight
point later ho there are infinite
circles Jo 8.8 through sec users a clear
Agoura will be doe pointless a ho-hum
nada cocky still there are infinite
circles that can pass through any two
given points lake in those the agar
aapke will be team non collinear
pointless a ho then there is one and
only one circle that can pass through
any given three non collinear points
so those hamachi cut the hey you're
following us along like hum always kept
they have that there is any doubt or
criminalize eval quick question han
please base eject bay without any
hesitation go ahead and send us a
comment and hum we'll be more than happy
to provide answers clarifications to you
so the static is presentation mail Hey
ugly will be certified milling is
optically shukriya Tanabata