6.11
Circles - Part 2:
6.11.1 Construction of chords
1. Construction of chord in a circle.
6.11.1
Problem 1: Construct
a chord of length 3cm in a circle of radius 2cm
|
Step |
Construction |
|
|
1 |
Draw a circle of radius 2cm with O as center. |
|
|
2 |
Mark any point P on the circle |
|
|
3 |
With P as center and 3cm as radius, draw an arc
to cut the circle at Q |
|
|
4 |
Join P and Q ( PQ is the chord of length 3cm) |
Note that this chord divides the
circle into 2 regions called segments (PSQ and PTQ).
PSQ is smaller in size as compared to
PTQ.
Note that the diameter POR (4cm) is the chord of maximum
length.
Definition: The
region bounded by the chord and an arc is called ‘segment’ of the circle.
Observations:
1. Diameter is the
chord having maximum length.
2. A chord cuts
the circle into 2 segments.
3. Diameter cuts
the circle into 2 equal segments (semi circles).
2. Finding distance
between chord and center of the circle.
6.11.1
Problem 2:
Construct a chord of length 4cm in a circle of radius 2.5cm and measure the distance
between the centre of the circle
and the chord.
|
Step |
Construction |
|
|
1 |
Draw a circle of radius 4cm with
O as center |
|
|
2 |
Mark any point P on the circle |
|
|
3 |
With P as center and 4cm as radius, draw an arc to
cut the circle at Q |
|
|
4 |
Join P and Q ( PQ is the chord of length 4cm) |
|
|
5 |
Bisect the line PQ to find its
mid point M. (With P and Q as centers, draw arcs of radius more than half the
length of PQ on both sides of PQ and let these arcs meet at R and S). Let RS
Meet PQ at M. |
|
|
6 |
|
Exercise: Construct few more chords of
length 4cm and measure their distance from the center.
Observations:
1. As the length
of the chord increases, the distance between the chord and the center decreases
and becomes 0 for the diameter.
2. As the length
of the chord decreases, the distance between the chord and the center increases
and becomes R (radius) finally.
3. Chords of equal
lengths are equidistant from the center.
|
No |
Figure |
Properties of
angle at circumference |
|
1 |
|
ASBA is a minor segment.
Minor arc(ASB) subtends
acute angle(ACB) at the circumference. |
|
2 |
|
ASBOA is a semi circle.
Semi circle subtends a right angle at the
circumference. |
|
3 |
|
ASBA is a major segment.
Major arc subtends obtuse angle at the
circumference. |
Have you observed race tracks in your
annual sports day? All of them are circular tracks with different radii?
Definition:
Circles having same center but different radii are called ‘concentric circles’.
A straight line which cuts the circle at 2 distinct points
is called ‘secant’.
|
No |
Figure |
Properties |
|
1 |
|
Circles having same center but different radii are
called ‘Concentric circles’. C1, C2 and C3 are 3 circles with different radii OA, |
|
2 |
|
Circles have same radii but different centers are
called ‘Congruent circles’. C1 and C2 are 2 circles
having same radii OA (OA= |
|
3 |
|
A straight line which cuts the circle at 2
distinct points is called ‘Secant’. AD is a straight line which cuts the circle at 2
points, B and C. ABCD is a Secant. |
|
4 |
|
A straight line which meets the circle at only
one point is called ‘Tangent’.
The point where the
line touches the circle is called ‘Point of
contact’(P) XY is a straight line which touches the circle at only one
point P. XPY is a tangent to the circle at P. |
Theorem:
The tangent at
any point of a circle and the radius through this point are perpendicular to
each other
(In the below mentioned figure, Prove that OP is
perpendicular to RS).
Hint :
Construction: Extend OP such that OP=PQ, Let SR be the
bisector of OQ.
Prove that the triangles OPS and QPS are congruent using
SSS postulate.
1. Construction of
tangent at a point on a circle
|
Step |
Construction |
|
|
1 |
Draw a circle of given radius
(2cm) with center as O |
|
|
2 |
Mark any point P on the circle |
|
|
3 |
Join OP (this becomes radius) |
|
|
4 |
Extend OP to Q such that OP=PQ (P
is midpoint of OQ) |
|
|
5 |
Construct a perpendicular line
to OQ at P (With O and Q as centers, draw arcs of radius more than half the
length of OQ on both sides of OQ and let these arcs meet at S and R) |
|
|
6 |
The line SPR is the tangent at P
to the circle |
Note: Since SP is perpendicular to OQ, 
OPS = 900
2. Construction of
tangent to a circle from an external point
6.11.1
Problem 3: Construct
a tangent to a circle of radius 2cm from a point 5cm away from the center.
|
Step |
Construction |
|
|
1 |
Draw a circle of given radius
(2cm) with center as O |
|
|
2 |
Mark any point P at a given distance (5cm) away
from the center O, Join OP |
|
|
3 |
Bisect the line OP (With O and P
as centers, draw arcs of radius more than half the length of OP on both sides
of OP and let these arcs meet at R and S) |
|
|
4 |
Let RS meet OP at M (Note that M
is mid point of OP) |
|
|
5 |
Draw arcs of radius = |
PX and PY are tangents to the circle from P.
Observe that from a
point within a circle we can not draw any tangent, from a point on a circle we can draw only one tangent and from an
external point we can draw only two tangents to a circle.
|
Let us
observe the figure on the right hand side. C1 is a
circle with O as center.C2 is a circle with P as center. G and H
are points on the circles C1 and C2 respectively. The
straight line AB touches the circle C1 at G and C2 at H. We
notice that AB is a tangent common to both the circles C1 and C2 and is
called common tangent. Similarly
XY is a common tangent to circles C1 and C2. In the
this figure we notice that centers of both the circles are on the same side
of tangent |
|
|
Definition: A tangent which is common to two
or more circles is called a ‘common tangent’. Let us
observe the figure on right hand side. C1 is a
circle with O as center. C2 is a circle with P as center. The two
circles touch at Y. XYZ is
a tangent common to both the circles at Y. In this
figure, we notice that centers of both the circles are on different sides of
the tangent. |
|
Though the two figures have common tangents, have you
observed the position of centers in both the cases?
Definition: If the centers of the circles lie
on the same side of the common tangent then that tangent is called ‘Direct common tangent’.
If the centers of the circles lie on different sides of the
common tangent then that tangent is called ‘Transverse common tangent’.
3. Construction of
Direct common tangents to 2 circles of equal radii whose centers are at a given
distance from each other
6.11.1Problem
4: Draw direct common tangents to 2
circles of radii 2cm whose centers are 5cm apart.
|
Step |
Construction |
|
|
1 |
Draw a line OP=5cm |
|
|
2 |
Draw 2 circles, C1 and C2 of radii 2cm with O and
P as centers |
|
|
3 |
Extend OP in both directions so that it cuts C1
at X and C2 at Y. Let OP cut C1 at R and C2 at S |
|
|
4 |
Bisect the line XR (with X and R as centers, draw arcs
of radius more than half the length of XR on both sides of XR). |
|
|
5 |
Similarly bisect the line SY |
|
|
6 |
Let RO cut the circle C1 at A and C. Similarly
let SP cut the circle C2 at B and D. |
AB and CD are the direct common tangents to the 2 circles.
4. Construction of
direct common tangent to two circles of different radii.
6.11.1
Problem 5:
Construct direct common tangents to 2 circles of radii 3cm and 2cm whose
centers are 6.5cm apart.
|
Step |
Construction |
|
|
1 |
Draw
the circle C1 of radius 3cm with A as center |
|
|
2 |
Let AB
= 6.5cm |
|
|
3 |
Draw
the circle C2 of radius 2cm with B as center |
|
|
4 |
Draw
the circle C3 of radius of 1cm (difference between radii of C1
and C2 =3-2) with A as center |
|
|
5 |
Bisect
the line AB to get its mid point M (with A and B as centers, draw arcs of
radius more than half the length of AB on either sides of AB to meet at C and
D). Join CD to cut AB at M |
|
|
6 |
With M
as center and AM as radius, draw an arc to cut C3 at X. BX is
tangent to C3 at X |
|
|
7 |
Produce
AX to meet C1 at P |
|
|
8 |
Draw a
line from B, parallel to AXP (use set squares). Let this line cut C2
at R |
|
|
9 |
Join
PR. PR is the direct common tangent to the given 2 circles |
5. Construction of
transverse common tangent to two circles.
6.11.1
Problem 6:
Construct transverse common tangent to circles of radii 2cm and 1 cm whose
centers are 5.5 cm apart.
|
Step |
Construction |
|
|
1 |
Draw
the circle C1 of radius 2cm with A as center |
|
|
2 |
Let AB
= 5.5cm |
|
|
3 |
Draw
the circle C2 of radius 1cm with B as center |
|
|
4 |
Draw
the circle C3 of radius 3cm (sum of radii of C1 and C2
=2+1) with A as center |
|
|
5 |
Bisect
the line AB to get its mid point M (with A and B as centers, draw arcs of
radius more than half the length of AB on either sides of AB to meet at C and
D). Join CD to cut AB at M |
|
|
6 |
With M
as center and AM as radius, draw an arc to cut C3 at X. BX is
tangent to C3 at X |
|
|
7 |
Join XA
to meet C1 at P |
|
|
8 |
Draw a
line from B parallel to APX (use set squares).Let this line cut C2
at R |
|
|
9 |
Join
PR. PR is the transverse common tangent to the given 2 circles |
Did you notice the
difference between the above two constructions? In the case of direct common
tangent, we draw a third circle with radius
equal to the as difference between radii
and in the case of transverse common tangent, we draw a third circle with
radius equal to the
sum of radii. Except for this difference all
other steps are same.
6.11 Summary of learning
|
No |
Points to remember |
|
1 |
From an
external point we can draw two tangents to a circle. |
|
2 |
The
radius drawn at the point of contact is perpendicular to the tangent |