10. CIRCLES (SCERT)

SEBA Class 10 Chapter 10. CIRCLES

Chapter 10. CIRCLES

Class 10 Maths Chapter 10. CIRCLES Exercise 10.1 Solutions

1. How many tangents can a circle have ?

Solution: A circle have infinitely many tangents .

2. Fill in the blanks :

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

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

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

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

Solution: (i) A tangent to a circle intersects it in points .

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

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

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

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 . Length PQ is :

(A) 12 cm (B) 13 cm

Solution: In , we have

(D) cm .

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: In Figure , O be a centre of the circle.

EF is given line , CD is a tangent and AB is a secant of the circle . So, line EF parallel to secant AB and also the line EF parallel to tangent CD .

Class 10 Maths Chapter 10. CIRCLES Exercise 10.2 Solutions

1.From a point Q, the length of the tangent to a circle is 24cm and the distance of Q from the centre is 25cm .The radius of the circle is

(A) 7cm (B) 12cm

Solution: In figure :

Here , OQ = 25 cm and QT = 24 cm

In OQT , we have

Answer: (A) 7 cm

2. In fig. 10.11,if TP and TQ are the two tangents to a circle with centre O so that then is equal to :

(A) 60° (B) 70°

Solution: In figure :

Here, and

Since, OPTQ is cyclic quadrilateral , We have

Answer : (B) 70°

3. If the tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then is equal to

(A) 50° (B) 60° (C) 70° (D) 80°

Solution: In figure :

Since ,

then,

In , we have

Answer : (A) 50°

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution: let AB and CD are two tangent touch at X and Y of the diameter XY of the circle with centre O .

To prove: .

Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .

and

So, and XY is a transversal .

Proved .

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre .

Solution: let AXB and CYD are two parallel tangents at the point X and Y of the circle with centre O .

To Prove : XY is a line passes through the centre of the circle .

Construction : join OX and OY . we draw .

Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .

and

and OX is a transversal .

and and OY is a transversal .

So, XOY is straight line .

is a line passes through the centre of the circle . Proved

6. The length of a tangent from a point A at distance 5cm from the centre of the circle is 4cm.Find the radius of the circle .

Solution: Here, OP = 5 cm and AP = 4 cm

In , We have

Therefore, the radius of the circle is 3 cm .

7. Two concentric circles are of radii 5cm and 3cm .Find the length of the chord of the larger circle which touches the smaller circle.

Solution: Here, OM = 3 cm and OP = 5 cm .

Solution: Here, OM = 3 cm and OA = 5 cm .

In , we have

Since,

Therefore, the length of the chord of the larger circle is 8 cm .

8. A quadrilateral ABCD is drawn to circumscribe a circle a circle (see Fig. 10.12). Prove that

Solution: let, be a quadrilateral and its sides are , ,and in which the points and are touch of a circle with centre O respectively .

To prove : .

Proof : Since the length of the two tangents from an external point to a circle are equal .

(From A)…….(i)

(From B)…….(ii)

(From C)…….(iii)

(From D )…….(iv)

Addin (i) ,(ii) ,(iii) and (iv) , we have

Proved.

9. In Fig. 10.13, XY and are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B .Prove that .

Solution: Given , and ' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting at A and ' at B . To Prove

Construction : Join and .

Proof : Since the lengths of the two tangents from an external point to a circle are equal and also the subtend equal angles at the centre .

Therefore , is a bisector of

and is a bisect of

Since, and is a transversal

[From and ]

In we have,

[ From ]

Proved.

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line –segment joining the points of contact at the centre.

Solution: Given, AP and BP be two tangents drawn from an external point P to a circle with centre O .

To prove :

Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .

Therefore , and .

So, and .

OAPB is cyclic quadrilateral we have

proved .

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution: Given , ABCD be a parallelogram and its sides AB , BC , CD and AD are touch the circle at P , Q , R and S respectively .

To prove : ABCD is a rhombus .

Proof : Since the lengths of tangents drawn from an external point to a circle are equal .

Therefore , (From ) …….

(From B )…….

(From C )…….

(From D )…….

Since, be a parallelogram . Thus, and

So ,

Therefore, is a rhombus.

12. A triangle ABC is drawn circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Solution: Since, the lengths of the two tangents from an external point to a circle are equal.

cm ; cm ; let, cm

Therefore, , ,

So,

A/Q,

(impossible)

Thus, cm and cm

13. Prove that the opposite sides of a quadrilateral circumscribing a circle subtended supplementary angles at the centre of the circle.

Solution: Given , ABCD is a quadrilateral circumscribing a circle with centre O and touches the quadrilateral at P , Q , R and S respectively .

To prove : (i)

(ii)

Construction : Join OP , OQ , OR and OS respectively .

Proof : Since the lengths of the two tangents from an external point to a circle are equal .

i.e., ; ; and .

In AOS and AOP , we have

(Given)

[ Common side]

[ R.H.S rule]

[ C.P.C.T]

Similarly , ; ;

Let, ,, , , , , and

We know that the sum of the all angles of subtended at a point is 360° .

Similarly , Proved.