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
(C) 8.5 cm (D) 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 .
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
(C) 15cm (D) 24.5cm
Solution: In figure :
Here , OQ = 25 cm and QT = 24 cm
In OQT , we have
cm
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°
(C) 80° (D) 90°
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, OP is the angle bisector of .
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
AX||OZ 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
cm
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 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 ABCD be a quadrilateral with sides AB, BC, CD, and AD, and let P, Q, R and S be the points where the circle with centre O touches the sides, 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 nagle 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: Let ABCD be a parallelogram with sides AB, BC, CD, and AD, and let P, Q, R and S be the points where the circle with center O touches the sides, 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)
or
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.