Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.1

Question 1.

Fill in the blanks :

(i) A tangent to a circle intersects it in __________ points

Solution:

one

(ii) A line intersecting a circle in two points is called a _________

Solution:

secant

(iii) A circle can have _________ parallel tangents at the most,

Solution:

two

(iv) The common point of a tangent to a circle and the circle is called _______

Solution:

point of contact

(v) We can draw __________ tangents to a given circle

Solution:

infinite

Question 2.

A tangent PQ of a point P’ of a circle of radius 5 cm meets a line through the centre ‘O’ at a point Q so that OQ = 12 cm. Find the length of PQ.

Solution:

PQ is a tangent to a circle whose centre is ‘O’.

P is the point of contact. PO is the radius of the circle.

Hence ∠OPQ = 90°

Given that OP = 5 cm and OQ = 12 cm

By pythagorus theorem.

(OP)^{2} + (PQ)^{2} = (OQ)^{2}

(PQ)^{2} = (OQ)^{2} – (OP)^{2}

(PQ)^{2} = 12^{2} – 5^{2} = 119

PQ = \(\sqrt{119}\)

Hence, length of PQ = \(\sqrt{119}\) cm

Question 3.

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution:

The adjacent figure is a circle with centre ‘O’

AB is the given line CD is a tangent to the circle of P parallel to AB.

EF is a secant of the circle, drawn parallel to AB

Question 4.

Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm. (A.P. Mar. ’16, 15)

Solution:

In ∆OTR ∠OTP = 90°

OT = 9 cm

OP = 15 cm

By pythagorus theorem

(OP)^{2} = (OT)^{2} + (PT)^{2}

⇒ 15^{2} = 9^{2} + (PT)^{2}

⇒ PT^{2} = 15^{2} – 92 = 225 – 81

(PT)^{2} = 144 ⇒ PT = \(\sqrt{144}\) = 12

Length of the required tangent = PT = 12 cm

Question 5.

Prove that the tangents to a circle at the end points of a diameter are parallel. (A.P. June ’15)

Solution:

PQ is the diameter of a circle with centre ‘O’. APB and CQD are tangents to the circle at P and Q respectively.

We have to prove that AB || CD

AB is a tangent to the circle ‘P’ is the point of contact and

PO is the radius drawn through P

∴ ∠APO = 90° ……………… (1)

CD is a tangent to the circle. Q is the point of contact and QO is the radius draw through Q.

∴ ∠DQO = 90°

From (1) and (2), we have

∠APO = ∠DQO

AB, CD are two lines. PQ is a transversal ∠APO and ∠DQO are a pair alternate angles. Since the alternate angles are equal AB || CD