Circles — Class 10 Mathematics
Mathematics · Chapter 10
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Chapter summary
Key ideas & terms
- Tangent
- A line that touches a circle at exactly one point.
- Point of contact
- The single point where a tangent meets the circle.
- Tangent–radius property
- A tangent is perpendicular to the radius at the point of contact.
- Equal tangents
- Tangents from an external point have equal length.
Extra questions & answers
What is the angle between a tangent and the radius at the point of contact?
How many tangents can be drawn from an external point, and what about their lengths?
Exactly two, and they are equal in length.
What is a tangent to a circle?
A line that touches the circle at exactly one point, called the point of contact (refer to the figure in your NCERT textbook).
How many tangents can be drawn to a circle from a point inside, on, and outside it?
None from a point inside, exactly one from a point on the circle, and exactly two from a point outside the circle.
State the theorem about a tangent and the radius at the point of contact.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
State the theorem about the lengths of two tangents from an external point.
The lengths of the two tangents drawn from an external point to a circle are equal.
A tangent PQ touches a circle of radius 5 cm at Q, and
Since
What is the angle between a tangent to a circle and the radius at the point of contact?
What is a secant of a circle?
A line that intersects the circle at two distinct points.
From an external point, two tangents of length 8 cm are drawn to a circle. What can you say about them?
They are equal in length (8 cm each), as tangents from an external point are always equal.
EXERCISE 10.1
1. How many tangents can a circle have?
A circle can have infinitely many tangents because exactly one tangent can be drawn at each point on the circle.
2. 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) A circle can have ______ parallel tangents at the most.
two
(iv) The common point of a tangent to a circle and the circle is called ______.
the point of contact
3. A tangent PQ at a point P of a circle of radius 5
textcm meets a line through the centre O at a point Q so that OQ=12
textcm. Length PQ is :
(A) 12
textcm (B) 13
textcm (C) 8.5
textcm (D)
sqrt119
textcm.
Since the radius OP is perpendicular to the tangent PQ, triangle OPQ is right-angled at P. Therefore,
Hence,
Therefore, the correct option is (D).
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.
Draw a circle with centre O and draw the given line l. Draw a radius OP perpendicular to l, where P lies on the circle. Through P, draw a line parallel to l; this line touches the circle only at P, so it is a tangent. Choose a point R inside the circle on OP and draw another line through R parallel to l. It intersects the circle at two points, so it is a secant.
EXERCISE 10.2
1. From a point Q, the length of the tangent to a circle is 24
textcm and the distance of Q from the centre is 25
textcm. The radius of the circle is
(A) 7
textcm (B) 12
textcm
(C) 15
textcm (D) 24.5
textcm
Let O be the centre and P the point of contact. Since OP
perpQP, triangle OQP is right-angled at P. Thus,
Therefore,
The correct option is (A).
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \angle POQ = 110^\circ, then
anglePTQ is equal to
(A) 60^\circ (B) 70^\circ
(C) 80^\circ (D) 90^\circ
The radii OP and OQ are perpendicular to the tangents TP and TQ. Hence, in quadrilateral OPTQ,
Therefore,
So,
The correct option is (B).
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80^\circ, then
anglePOA is equal to
(A) 50^\circ (B) 60^\circ
(C) 70^\circ (D) 80^\circ
Since PA=PB, OA=OB, and OP is common, triangles OAP and OBP are congruent. Thus, OP bisects
angleAPB, giving
Also, OA
perpPA, so \angle OAP=90^\circ. In triangle OAP,
The correct option is (A).
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB be a diameter of a circle, and let the tangents at A and B be l and m, respectively. By the tangent-radius theorem,
and
Since OA and OB lie on the same straight line AB, both l and m are perpendicular to AB. Two lines perpendicular to the same line are parallel. Therefore,
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Let a tangent touch a circle at P, and let O be the centre. The radius OP is perpendicular to the tangent at P. Through a given point on a line, only one perpendicular can be drawn to that line. Therefore, the perpendicular to the tangent at P is the line OP, which passes through the centre O.
6. The length of a tangent from a point A at distance 5
textcm from the centre of the circle is 4
textcm. Find the radius of the circle.
Let O be the centre and P the point of contact. Since OP
perpAP, triangle OAP is right-angled at P. If the radius is r, then
Thus,
Hence,
7. Two concentric circles are of radii 5
textcm and 3
textcm. Find the length of the chord of the larger circle which touches the smaller circle.
Let AB be the chord of the larger circle touching the smaller circle at P, and let O be their common centre. Since AB is tangent to the smaller circle,
The perpendicular from the centre bisects the chord, so AP=PB. In right triangle OAP,
Thus,
Therefore,
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB+CD=AD+BC
Let the circle touch AB, BC, CD, and DA at P, Q, R, and S, respectively. Tangents drawn from the same external point are equal, so
Now,
Using the equal tangent lengths,
Therefore,
9. In Fig. 10.13, XY and X
′
Y
′
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 \angle AOB = 90^\circ.
Let P and Q be the points of contact of the parallel tangents XY and X
′
Y
′
, respectively. From the external point A, the two tangents are AP and AC. Therefore, AO bisects the angle between them:
Similarly, from the external point B, the tangents are BQ and BC, so
Since XY
parallelX
′
Y
′
and AB is a transversal,
Hence,
In triangle AOB,
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.
Let PA and PB be tangents from an external point P to a circle with centre O, where A and B are the points of contact. Since a radius is perpendicular to the tangent at the point of contact,
In quadrilateral OAPB, the sum of the interior angles is 360^\circ. Therefore,
Thus,
Hence, the angle between the tangents is supplementary to the angle subtended by the segment joining the points of contact at the centre.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Let ABCD be a parallelogram circumscribing a circle. For any quadrilateral circumscribing a circle,
Since ABCD is a parallelogram,
Therefore,
so
and hence
Thus adjacent sides of the parallelogram are equal. Since opposite sides are also equal, all four sides are equal. Therefore, ABCD is a rhombus.
12. A triangle ABC is drawn to circumscribe a circle of radius 4
textcm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8
textcm and 6
textcm respectively (see Fig. 10.14). Find the sides AB and AC.
Let the circle touch AB and AC at F and E, respectively. Tangents from the same external point are equal, so
Let
Then
The semiperimeter is
Since the inradius is 4
textcm, the area is
By Heron's formula,
Therefore,
Squaring both sides gives
Since x+14>0,
so
Hence,
and
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Let ABCD be a quadrilateral circumscribing a circle with centre O. Since O is equidistant from each pair of adjacent sides, OA, OB, OC, and OD bisect the angles A, B, C, and D, respectively. Therefore, in triangle AOB,
Similarly, in triangle COD,
Adding,
The angle sum of a quadrilateral is 360^\circ, so
Thus, the opposite sides AB and CD subtend supplementary angles at O. In the same way,
Hence, both pairs of opposite sides subtend supplementary angles at the centre.