Circle

Solution of NCERT Exercise 10.1

Question (1) How many tangents can a circle have?

Answer: Infinitely many

Explanation: Since the circumference of a circle can have infinitely many points, thus a circle can have infinitely many tangents.

Question (2) Fill in the blanks

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

Answer: A tangent to a circle intersects it in one point (s).

Explanation: A line that intersects the circle at only one point is called A Tangent. This means a tangent to a circle intersects it in only one point.

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

Answer: A line intersecting a circle in two points is called asecant.

(iii) A circle can have _______ parallel tangents at the most.

Answer: A circle can have two parallel tangents at the most.

Explanation: Because a circle can have only one point just opposite to any other point. Consequently a circle can have two parallel tangents at the most.

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

Answer: The common point of a tangent to a circle and the circle is called Point of contact.

Question (3) 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. Length PQ is:

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) `sqrt(119)`

Answer: (D) `sqrt(119)`

Explanation:

Let O is the centre of the given circle.

PQ is tangent and OP is the given radius

According to question, OP = 5 cm

OQ = 12 cm

PQ=?

Now, since OP is the radius and PQ is the tangent.

This means, OP is perpendicular to PQ

∴ ∠ OPQ = 90⁰

Now, according to Pythagoras theorem,

OQ² = OP² + PQ²

⇒ 12² = 5² + PQ²

⇒ 144 = 25 + PQ²

⇒ 144 – 25 = PQ²

⇒ 119 = PQ²

`:. PQ = sqrt(119) cm` Answer

Question (4) 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:

Draw a circle having centre O.

And draw a line AB near to the circle.

Draw a perpendicular NO on line AB which meets the centre O of the circle.

Now, draw a perpendicular EF on point P. This EF is the tangent to the circle having centre O.

Now, take another point M on ON.

Draw a perpendicular GH on point M

Here, Line EF and GH are parallel to line near AB

Here, EF is tangent to the circle having center O and parallel to line AB

And GH is the secant of the circle having center O and parallel to the line AB

MCQs Test

Back to 10th-Math-home

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