L5 : Circle through 3 points- Circles , Maths, Class 9 Video Lecture - Class 9 | Best Video for Class 9

hello friends this video encircles part
5 is brought to you by exam Viacom no
more fear from exam now let's try to
draw various circles from a given point
X suppose you have been given a point
one this is a point red dot point let's
try to draw it yourself as passing
through this point so we can draw
something like this like this like this
so you see there are so many circles
which I could draw from one point
infinite number of circles let's I see
with two months so we can draw one
circle like this like this like this if
you see from two point also we can draw
in finite circles in five surface Hey so
if you haven't given two points we can
draw in finite number because using
these two points let's try from three
points three point they are two option
one has the three points are the same
line
they are cooling here and they are non
collinear so let me write here and let's
try to draw a circle you'll see that you
can draw only one circle from three non
collinear points and from three quality
of points you can't draw a circle even
if you try and you see Third Point is
not moving here
8:09 from 300 Aranea points circle is
impossible and you can't draw only one
circle wrong from precoding a point you
cannot draw a circle that is our
observation is all activity you're doing
actually on the piece of paper okay so
the test there is if you tell them that
there is only one and only one circle
passing through
given three know on who will be advanced
please don't
three not will be eventually we have
three non put any points from that you
can have only one circle to prove it
so we have been given three point let's
try to move it let us join these points
okay and let us name this ba we see now
let's draw the perpendicular bisector
for these lines a B and E Z okay I just
drew a perpendicular bisector for these
lines you can actually do there are so
many ways of doing it active in using to
ease in protector or you can actually
means that you see scale at this point
and at this point you draw perpendicular
whatever ways you want you can do it
okay I'm assuming is a perpendicular
bisector for these and then this might
need that now if you see this point or
will actually lies on the perpendicular
bisector of a B and O will also my you
the perpendicular bisector of VZ okay
now let's do one thing let's join these
points for a movie you know see and
showing this much that's very smart
now you know observe that away will be
quite 200 is the movie why because if
you see this is the perpendicular sector
so you can actually prove that these two
triangles are this so triangle a om and
triangle bo and you see that both are
confident why this common side is the 90
degree and this this side is the same so
using the congruence of triangles you
can actually do that with the sides are
equal right so Oh a will be equal to
movie similarly if you take these two
triangles where we
CEO ends but Lionel you see see you can
prove that Oh seasonal rule you can
actually do that you can do that East we
should like us a comment oh it's going
to be through these triangles are
congruent to B is equal to C do you
think these two equation if it miles we
made what that means all these three
lengths are equal in this length are
equal we can actually consider them to
be a radius and draw a circle because
these three lengths are equal
Oh a movie you know see I just will call
it an Jamie's and thus we can actually
draw one circle you see when ABC are
actually equal distance from this point
oh okay so if you draw a circle from the
point O with a length of a or a or OB or
a ceases or are same it is an you'll
actually draw a circle and this circle
will pass through all these points a B
and C and now since these two
perpendicular bisector which we have
drawn green can actually intersect only
at one point right because these two
lines in e29 get intersect only at one
point thus they do only one Center since
two lines in this case perpendicular
bisector of a B and V can intersect only
at one point there will be only one
Center please no important concept here
lie some brain we have to prove that
there is only one C using the first
thing we prove that we can draw one
circle but what is the guarantee that
you can draw only one circle for that we
have to prove that see any two line if
you pay it can meet only at one point
similarly here these two perpendicular
bisector
we'll only meet at one time sent they'll
meet only at one point they can be only
one Center for a circle since there is
only one center for a circle
hey that means there is only one circle
okay there's six immunocal here so it
has given that we have implement an arc
and be able to complete a circle so what
we can do is you can actually take three
points in the circle on this arc now we
have three points and now we also know
how to draw a circle of a three point we
can apply the same logic okay so that it
is as good as we had three points draw a
circle
Hey so if we are given an odd this plot
three points of the arc and for it to
put the arc now using three points for
the circle pretty easy join these points
so once we have joined this point and at
this point speed ABC so once you join a
V and easy we can draw a perpendicular
bisector of a b and b c and let that
meet at point O so in the speed at point
O what we can do is we can draw a circle
from point O consider 8 OC will be all
the way as now the radius Buddhist all
are saying so we have seen this okay
next question is draw different pairs of
circle in how many points does each pair
having guards that means we have to draw
pay off circle it will be a road or two
circles and we have to see how many
points that's each pair have in common
what is the maximum amount let's take
these two circle then let's try to move
the green circle okay so now you see it
at this state at this state how many
common points 0 good idea so at this
state there is a 0 common fine
let's move this green circle ahead so
now if you see how many common points
two columns
9 to convert its mark this cone find
here and receive two more words
there's move the circle ahead but that
one you see that there's only one
and this is the contract hey hey Circle
ahead you see that again theta0 bond
which we have seen for snap use okay
miss sweet once again zero common point
that we have seen two common points one
common point against either condom so if
you want to draw this pair of circle you
get draw these all these four scenarios
we discuss and you can you say that the
maximum number of common point is so the
key takeaway from this cushion is any to
circle the maximum number of common
point did we do range we have two
circles like this this is zero power
point should be able to circle just
touching each other this is one common
point to circle like this there are two
common why it's two circle one aside to
the zero common point right 0 1 2 n
vaccine common point can be to the
minimum can be 0 and we can also have a
scenario where there is only one common
point between two cells suppose you are
given a circle give a construction to
find its center so here we have we are
given the circle and we have to find a
center same thing again this take three
points so you have three points for the
circle then name is a B and C we can
join these points and then we can draw a
perpendicular bisector of a B and VC and
the perpendicular bisector black center
of the circle and this point is called
non center of the circles finding center
is pretty easy so if you have a circle
this property quantum circle and then
join these points a B and VC draw the
perpendicular bisector of a B and easy
and this perpendicular bisector lacking
with the point that point will be the
center of a circle by because we have
seen that actually Oh a OB no CR or the
same length and that length is called
phase if two circles intersect at two
points to that they are Center life on
the perpendicular bisector of the common
God so there to service and they
intersect at two points you see here two
points here
okay and we have seen that the only way
they can intersect at this point is this
possible scenario here to prove that
they sit there lying on the
perpendicular bisector of the common
chords first draw the common god and let
me name it he ain't bdbb the common God
and you will know that the same test
lies on the perpendicular bisector of
account so let's draw a perpendicular
bisector of the common God
so here this length is equal to this
length and this is 98 this is 90 thank
so we held it this point we see so what
is given given AC is equal to VZ first
thing it is also given then let this
point be over - and given that
these two things are given okay so we
have to prove that oh and would ask and
see all these liners in line with the
particulars it is nothing but this whole
line and let it let me name index by all
these XY is the particular bisector so
if you prove that
oh I know - by you this X see this is
the logical it's believed he means the
core of the circle the center hole right
so a be the same chord of the circle oh
that means perpendicular bisector of av
this is X C will actually pass through
whoa hey so I can say that OC exists
similarly a B is also the cord for the
center of a circle would ask the the
perpendicular bisector of a B here that
is o - see they will pass through what
has particular bisector this pass
through that means what is the exist
correct that's mean I can see that the
center of the circle actually lies on
the perpendicular bisector that person
per meter bisector is XY see what I can
say is that since he is called for
circle with Center oh okay so
perpendicular bisector by sit down in
this case it is XY okay Oh will your
next way because the center will
actually buy you on that's perpendicular
method similarly since a B is chord for
circle with Center would ask
so again perpendicular bisector XY to
take this would ask you like on this
right so oh well my on XY would hassle
on you next time
that means both oh I notice with my own
expand and explain nothing but the
perpendicular bisector okay okay but if
you go bisector even if this evening if
you have this ami have only one company
to bisector so in both cases you have a
common chord so we have a common
perpendicular bisector thank you visit
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