Question 11
A toy is made of a circle of radius 4 cm and a triangle as shown in figure.
If pictorially it is represented as triangle ABC such that the segments BD and DC into which
BC is divided by the point of contact, are of lengths 8 cm and 6 cm respectively. Find the
length of sides AB and AC of the toy
Solution
Either x+14=0 or x-7 =0
i.e., x= -14 (not possible) or x= 7
So, AB = x+8 = 7+ 8 =15 cm
CA = 6+x = 6+7 = 13cm
Question 12
Susma wanted to make a decor item with a bangle and thread. After tying the thread through
the diameter and another point on the circle as shown in figure, she realized the situation can
be expressed geometrically. So she drew the adjacent figure to represent it. She thought of
the thread as tangent at a point C of a circle and diameter as AB which when extended
intersect the thread from point C, at P. On measuring she found ∠PCA=110º help her find
∠CBA.
Solution
The required figure for the problem
situation is represented by the fig.
Join OC. As OC is the radius.
Since, tangent at any point of a circle
is perpendicular to the radius through
the point of contact.
∴ OC ⊥ PC
Now, ∠PCA = 110° [Given]
⇒ ∠PCO + ∠OCA = 110°
⇒ 90° + ∠OCA = 110°
⇒ ∠OCA = 20°
∵ OC = OA = radius of circle
∴
∠OCA =
∠OAC = 20°
[Sides opposite to equal angles are equal] ⇒ 50° + 20° + ∠PBC = 180° ⇒ ∠PBC = 180° – 70° ⇒ ∠PBC = 110°
Since, ABP is a straight line. ∴ ∠PBC + ∠CBA = 180° ⇒ ∠CBA = 180° – 110° = 70°
Since, PC is a tangent, ∠BCP = ∠CAB = 20° [Angles in alternate segment]
In ∆PAC, ∠P + ∠C +
∠A = 180°
∠P = 180°
–
(
∠C +
∠A)
= 180°
-(110°+ 20°)
= 180°
– 130° = 50°
In ∆PBC, ∠BPC + ∠PCB +
∠CBP = 180°
⇒ 50° + 20° +
∠PBC = 180°
⇒
∠PBC = 180°
– 70°
⇒
∠PBC = 110°
Since, ABP is a straight line. ∴ ∠PBC + ∠CBA = 180° ⇒ ∠CBA = 180° – 110°
= 70°
Question 13
In the figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Solution
AB = 2 (10/3)= 20/3 cm or 6.67 cm
Question 14
Two tangents TP and TQ are drawn to a circle with centre O from an external
point T. Prove that ∠PTQ = 2∠OPQ.
Solution
Prove ∠PTQ = 2 ∠OPQ
Question 15
Prove that the line segment joining the points of contact of two parallel tangents of a circle,
passes through its centre
Solution
CD and EF are two C parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is diameter of the circle.
Const.: Join OA and OB. Draw OM || CD.
Proof: ∠1 = 90° … (i)
…[∵ Tangent is I to the radius through the point of contact
OM || CD
∴ ∠1 + ∠2 = 180° …(Co-interior angles
90° + ∠2 = 180° …[From (i)
∠2 = 180° – 90o = 90°
Similarly, ∠3 = 90°
∠2 + ∠3 = 90° + 90° = 180°
∴ AOB is a straight line.
Hence AOB is a diameter of the circle with centre O.
∴ AB passes through centre 0
Question 16
∆ABC, AB = 8 cm, BC = 6 cm, CA = 4 cm. With the vertices of triangle as centre, three
circles are described, each touching the other two externally. Find the radii of each circle
Solution
x+ y = 6 cm …(1)
y + z = 4 cm …(2)
z + x = 8 cm …(3)
Adding (1), (2), (3), we get
2 (x + y + z) = 18
x + y + z = 9 …(4)
(4) – (1) gives, z = 3
(4) – (2) gives, x = 5
(4) – (3) gives, y = 1
∴ Radii of circles are 5 cm, 1 cm and 3
SELF ASSESSMENT
Q1. Prove that the parallelogram circumscribing a circle is a rhombus.
Q2. Prove that the tangent at any point of a circle is perpendicular to the radius through the point
of contact
Q3. Prove that the lengths of tangents drawn from an external point to a circle are equal.
Q4. TP And TQ are tangents from T to the circle with Centre O and R is any point on circle. If
AB is the tangent to circle at R then prove that TA+AR=TB+BR
Q5. Prove that the parallelogram circumscribing a circle is a rhombus.
Q6. Two concentric circles are of radii 5 cm and 3 cm,find the length of the chord of the larger
circle which touches the smaller circle
Q7. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary
angles at the centre of the circle.
Q8. Prove that the parallelogram circumscribing a circle is a rhombus.
