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. ACB is angle at the circumference and is formed by the
minor segment ASBA. Minor arc(ASB) subtends acute
angle(ACB) at the circumference. |
2 |
|
ASBOA is a semi circle. ACB is angle at the circumference and is formed by the semi
circle ASBOA. Note that ACB = 900. Semi circle subtends a right angle
at the circumference. |
3 |
|
ASBA
is a major segment. ACB is angle at the circumference and is formed by the major
segment ASBA. 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 |