Theorem on Lengths of Tangents Video Lecture | Crash Course: Class 10

hello friends welcome to rising pearl
calm today were talking about series 10
where we're learning a trade Elon
circles this is episode number 11 and
today's topic friends is that the
lengths of the tangents drawn from an
external point to a circle are of same
lengths so friends in the last episode
we certainly saw that from an external
point which is a point outside a circle
you can draw only two tangents to that
circle so in this very sort the theorem
this is actually a theorem and we should
also add that this is the second very
very very important theorem so there was
a first year that we saw I think two
webisodes ago and this is the second and
the final theorem that we are going to
learn for circles and tangents so this
theorem tells us that that the lengths
of the two tangents drawn from an
external point to a circle those two
tangents are of same length so friends
like me always so let's first find out
what is the fear of telling us so here
we have drawn ourselves a circle and
let's say the center is o now what we
are going to do is we are going to take
an external point let's say this
external point is P now we are going to
draw the two tangents so this is tangent
one this is tangent two and these two
tangents they touch the circle at two
fixed points so let's call this point as
a and let's call this point as B so from
one external point to a circle they can
draw only two tangents so PA and PB so
the theorem is telling us that the
length of this must be the same in other
words PA must be equal to PB this is
what the theorem is telling us like I
said this is a very very important
theorem and as we start to solve
questions on circles and tangents you
will make use of this almost always
let's find out now how do we prove this
so here we have our circle with centre o
and this is the external point P and
let's say this is point a and
point B so what we are going to do is be
able to first joined Oh a and O B and
then o and P we are going to join these
lines Oh a OB and open now since I am
going to show you two ways of proving
this theorem so first is the more
conventional way which is that if you
compare this triangle and this triangle
now this angle is 90 degrees and this
angle is 90 degrees so why are these
angles 90 degrees the reason these two
angles are 90 degrees is that now PA is
tangent to circle with Center o the PA
is tangent now at the point where the PA
touches the circle oh we have drawn a
radius Oh a so from the first theorem if
this is our radius o is a radius which
is drawn from the point where PA is
tangent then the radius and the tangent
they are perpendicular to each other so
because of that we can say Oh a P is 90
degrees for the exact same reason we
will say Oh BP is also equal to 90
degrees because P P is a tangent and OB
is a radius this is 90 degrees now Oh a
is equal to OB
both our radius of the same circle so
these two are the radius of the same
circle so they are both of the same
length now this length is common
Oh P is common to both the triangles
this is a common side and this is the
hypotenuse right because the side
opposite to 90 degree this is 90 degree
so the side opposite to 90 degree is the
hypotenuse so this is the hypotenuse and
this is actually common to both the
sides so we can say that then we can
actually write this as triangle Oh a P
this triangle and triangle
Oh b.p these two triangles are actually
congruent triangles we have seen what is
the meaning of congruent triangles
these two triangles are congruent by the
theorem our H s which is a right angle
hypotenuse side congruence theorem as
per that these two triangles are
congruent by RHS theorem now we know
that if these two triangles are
congruent they are identical and
corresponding parts of congruent
triangles are equal so because of that
this length must be equal to this length
and so PA must be equal to PB and hence
it is proved that the two tangents
lengths of two tangents PA and PB are
the same now what about the second
method the second method goes something
like this so if you look at triangle o
AP this is a right angle triangle right
so ap we can from Pythagoras theorem we
can say that ap squared ap square plus R
square this is open let's say this
radius is our R square is equal to o P
squared because this is a right triangle
o AP and this is the hypotenuse so as
the Pythagoras theorem R square plus ap
square is equal to o P squared so this
is equal to o P Square or from here we
can say that ap square is equal to o P
square minus R square now if you look at
this right triangle Oh beeping so this
is again a right triangle so from here
we can say PD square or DP square be P
Square + R square which is this one R
squared is equal to o P Square the
hypotenuse or from here you can say that
be P Square
is equal to Opie square minus R square
so what you will notice is that we don't
know what is the value of Opie but ap
square is equal to OB square minus R and
B P squared is also equal to o P Square
minus R so even though we don't know
what is the value of R and ope but ap
square is this value and BP squared is
this value so we can say both the
squares are the same
it means AP squared is equal to VP
square or ap is equal to BP so this is
one other way in which you can prove it
so thanks for the rest of the baby so we
are going to try to write the proof
itself in a nice clear by the way you
would want to write it in the exam and
we will use for the purposes of this
very so we are going to use this
technique the more conventional
technique so now first we are going to
write what is given so we we have a
circle with center o P is an external
point we draw two tangents from P to the
circle so let's first draw our circle so
here we have a circle with centre o we
have the point P which is the external
point and PA and PB are the two tangents
this is what we have now we write what
is that we have to prove we have to
prove that PA is equal to PB this is
what we have to prove then we write all
the construction so the first cultural
construction is joined o n a so we do
that then we join o and B onb and also
oh and P or P & Co which will be like
this so now that we have finished our
construction now what we will see is
that let's now begin the proof itself so
if we look at let's write it so P a B
and this is a center o if you consider
all ap and OB p o AP and o BP these two
triangles
if you look at these two triangles what
do we see we know that Oh a equal to OB
so these two lengths are equal because
both are radius of the same circle so
they should be equal next Opie is a
common side so that we see that Opie is
common and angle OAP equals angle OBP
and both are 90 degrees this is 90
degrees and this is 90 degrees now why
are these two angles 90 degrees because
we know that the radius and the tangent
are perpendicular to each other so both
of them are 90 degrees so we say that
these two triangles are congruent and
they are corn fluid by RHS confluence
rule and hence because we know
corresponding sites in congruent
triangles are equal we can write P a
equals P B and hence it is proved that
the length of the two tangents drawn
from an external point both these
lengths are the same