## AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

### 10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.

Fill in the blanks.

i) A tangent to a circle intersects it in ——— point(s). (one)

ii) A line intersecting a circle in two points is called a ———. (secant)

iii) The number of tangents drawn at the end of the diameter is ———. (two)

iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)

v) We can draw ——— tangents to a given circle. (infinite)

Question 2.

A tangent PQ at 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 length of PQ.

Answer:

Given: A circle with centre O and radius OP = 5 cm

\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm

We know that ∠OPQ = 90°

Hence in △OPQ

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

[∵ hypotenuse^{2} = Adj. side^{2} + Opp. side^{2}]

12^{2} = 5^{2} + PQ^{2}

∴ PQ^{2} = 144 – 25 .

PQ^{2} = 119

PQ = √119

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.

Answer:

Steps:

- Draw a circle with some radius.
- Draw a chord of the circle.
- Draw a line parallel to the chord intersecting the circle at two distinct points.
- This is secant of the circle (l).
- Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.

Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.

Answer:

Given: A circle with radius OP = 9 cm

A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm

Now in △POQ, ∠P = 90°

OP^{2} + PQ^{2} – OQ^{2}

9^{2} + PQ^{2} = 15^{2}

PQ^{2} = 15^{2} – 9^{2}

PQ^{2} = 144

∴ PQ = √144 = 12 cm.

Hence the length of the tangent =12 cm.

Question 5.

Prove that the tangents to a circle at the end points of a diameter are parallel.

Answer:

A circle with a diameter AB.

PQ is a tangent drawn at A and RS is a tangent drawn at B.

R.T.P: PQ || RS.

Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.

∴ OA ⊥ PQ ……….(1)

[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]

Similarly, OB ⊥ RS ……….(2)

[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]

But, OA and OB are the parts of AB.

i.e., AB ⊥ PQ and AB ⊥ RS.

∴ PQ || RS.

O is the centre, PQ is a tangent drawn at A.

∠OAQ = 90°

Similarly, ∠OBS = 90°

∠OAQ + ∠OBS = 90° + 90° = 180°

∴ PQ || RS.

[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]