CIRCLES AND CONSTRUCTIONS

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CIRCLES AND CONSTRUCTIONS
IMPORTANT TERMS
CIRCLES
1. Circle: A circle is a collection of all points in a plane which are at a constant distance
[radius] from a fixed point [centre].
2. Tangent: A tangent to a circle is a line that touches the circle at exactly one point.
3. Point of contact: The point at which the line meets the circle is called its point of
contact and the tangent is said to touch the circle at this point.
4. Secant: A line which intersects the circle at two points is called a secant of the circle.
5. Length of tangent: The length of the segment of the tangent from the external point
and the point of contact with the circle is called the length of the tangent from the
external point to the circle.
TANGENTS AND THEIR PROPERTIES
1. There is no tangent to a circle passing through a point lying inside the circle.
2. There is one and only one tangent to a circle passing through a point lying on the circle.
3. There are exactly two tangents to a circle through a point lying outside the circle.
THEOREM 1: The tangent at any point of a circle is perpendicular to the radius
through the point of contact.
Given: A circle with centre O and a tangent AB at a
point P of the circle.
To prove: OP ⊥ AB
Construction: Take a point Q on AB. Join OQ.
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O
R
A P Q B
Proof: Q is a point on the tangent AB, other than the point of contact P.
For example:
Q lies outside the circle.
Let OQ intersect the circle at R.
Then, OR < OQ
..... (i)
But, OP = OR [radii of the same circle]. ..... (ii)
From (i) and (ii)
OP < OQ
OP is the shortest distance between the point O and the line AB.
But, the shortest distance between a point and a line is the perpendicular
distance.
OP ⊥ AB.

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In the figure O is the centre of a circle; PA and PB is
the pair of tangents drawn to the circle from point P
outside the circle;
If AOB = 117º, find .
In quadrilateral AOBP, A = B = 90º
[Radius is ⊥ to tangent at the point of contact]
A
P 117º
AOB + APB = 180º [ sum of angles of quad. is 360º]
117º + = 180º
= 180º – 117º = 63º
= 63º
B
O
THEOREM 2: The lengths of tangents drawn from an external point to a circle
are equal.
Given: Two tangents AP and AQ are drawn from a point A to a circle with centre
O.
To prove: AP = AQ
Construction: Join OP, OQ and OA.
Proof: AP is a tangent at P and OP is the
radius
through P.
OP ⊥ AP
Similarly, OQ ⊥ AQ
In the right OPA and OQA, we have
OP = OQ
OA = OA
A
[radii of the same circle]
[common]
OPA = OQA [both 90º]
OPA OQA [by RHS-congruence]
Hence, AP = AQ.
[cpct]
For example:
In figure PA and PB are tangents to the circle drawn
from an external point P. CD is another tangent
touching the circle at Q. If PA = PB = 12 cm and QD =
3 cm, find the length of PD.
Point D outside the circle. DQ and DB is the pair of
tangents drawn to the circle from the point D.
DB = DQ
DB = 3 cm
( QD = DQ = 3 cm is given)
Now, PD + DB = PB
PD + 3 cm = 12 cm ( PB = 12 cm is given)
PD = 9 cm
For example:
A quadrilateral ABCD is drawn so that D = 90º, BC = 38 cm and CD = 25 cm. A circle
is inscribed in the quadrilateral and it touches the sides AB, BC, CD and DA at P, Q,
R and S respectively. If BP = 27 cm, find the radius of the inscribed circle.
P
C
D
Q
P
Q
A
B
O
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Let O be the centre and r be the radius of the
inscribed circle. Join OR and OS.
Here, D = 90º
ORD = OSD = 90º
OR = OS
Therefore, ORDS is a square.
RD = SD = r ..... (i)
[1]
BP and BQ are tangents to the circle from point P.
BQ = BP
BQ = 27 cm
D
S
A
25 cm
R
C
r Q
r O
P
27 cm
38 cm
B
Then CQ = BC – BQ = 38 cm – 27 cm.
CQ = 11 cm
CR = 11 cm ..... (ii)
Now, RD + CR = CD
r + 11 cm = 25 cm By (i) and (ii)
r = 14 cm.
GEOMETRICAL CONSTRUCTIONS
1. The perpendicular bisector of a chord passes through the centre.
2. The tangent of a circle is perpendicular to the line joining the centre and the point of
contact.
3. The similar triangles or quadrilaterals have their corresponding angles equal and
corresponding sides in same ratio.
4. The point of concurrence of bisectors of angles of a triangle is called the incentre of
the circle inscribed in the triangle.
5. The point of concurrence of perpendicular bisectors of a triangle is called the
circumcentre of the circle circumscribed through the vertices of the triangle.
6. The lengths of two tangents drawn from an external point to a circle are equal.
For Example:
Construct one isosceles triangle whose base is 8 cm and altitude 4 cm and then
1
construct another triangle whose sides are 1 times the corresponding sides of the
2
isosceles triangle.
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Steps of Construction:
1. Take BC = 8 cm.
2. Draw DA, the right-bisector of BC and
equal to 4 cm.
3. Join BA and CA. ABC is the isosceles
triangle of given data.
4. Let-any acute angle be CBX (< 90).
5. Mark the three points B 1, B 2 and B 3 such
that BB 1 = B 1B 2 = B 2B 3·
B
B 1
D
A
B 2
A
8 cm
6. Joint B 2C. Draw a line through B 3 parallel to B 2C which meets BC extended at C'.
7. Through C' draw C'A' parallel to CA which meets BA produced at A'.
Then A'BC is the required triangle whose sides are
of the given triangle
For example:
3
2
4 cm
B 3
times the corresponding sides
C
X
C
X
Draw a line segment AB of length 6 cm. Taking A as centre, draw a circle of radius 3
cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to
each circle from the centre of the other circle.
Steps of construction:
1. Draw a line segment AB = 6 cm.
2. With centre A and radius = 3 cm, draw a circle.
3. With centre B and radius = 2 cm, draw
another circle.
4. With M, the midpoint of AB, as centre and
radius AM = MB, draw the circle
intersecting the circle of radius 3 cm at R
and S and intersecting the circle of radius
2 cm at P and Q.
5. Join AP, AQ, BR and BS.
Then AP and AQ are tangents to the circle with
centre B and radius 2 cm from point A
BR and BS are tangents to the circle with centre A and radius 3 cm from point B.
A
R
S
3 cm M 2 cm
P
Q
B
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