1. How many tangents can a circle have?
1. How many tangents can a circle have?
Explanation:
There can be many tangents to a circle. An infinite number of points that are all equally spaced from one another make up a circle. A circle's circumference contains an infinite number of points, allowing for the creation of an infinite number of tangents.
2. Fill in the blanks:
(i) A tangent to a circle intersects it in _______________ point(s).
(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 _______________.
2. Fill in the blanks:
(i) A tangent to a circle intersects it in _______________ point(s).
(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 _______________.
Explanation:
(i) A tangent to a circle intersects it at one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) 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 the point of contact.
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) cm
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) cm
Explanation:
(D) The radius along the point of contact is given by OP, while PQ represents the tangent.
OPQ = [Any point on a circle has a tangent to the radius passing through the point of contact.]
In right angled triangle OPQ,
OQ2 = OP2 + PQ2 [By Pythagoras theorem]
= 144 – 25 = 119
PQ = 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.
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.
Explanation:
Parallel lines XY and AB in the image above. Line segment AB is the tangent at point C, while line segment XY is the secant.
5. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
5. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Explanation:
(A) OPQ =
[Any point on a circle has a tangent to the radius passing through the point of contact]
In right angled triangle OPQ,
[By Pythagoras theorem]
= 625 – 576 = 49
OP = 7 cm
6. In the figure, if TP and TQ are the two tangents to a circle with centre O so that POQ = then PTQ is equal to:
(A)
(B)
(C)
(D)
6. In the figure, if TP and TQ are the two tangents to a circle with centre O so that POQ = then PTQ is equal to:
(A)
(B)
(C)
(D)
Explanation:
(B) POQ = , OPT = and OQT =
[Any point on a circle has a tangent to the radius passing through the point of contact]
As per the OPTQ quadrilateral,
POQ + OPT + OQT + PTQ =
[using Angle sum property of quadrilateral]
+ PTQ =
+ PTQ =
PTQ =
7. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of then POA is equal to:
(A)
(B)
(C)
(D)
7. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of then POA is equal to:
(A)
(B)
(C)
(D)
Explanation:
(A) OPQ =
[Any point on a circle has a tangent to the radius passing through the point of contact]
OPA = BPA
[Centre will fall on the bisector of the angle between the two tangents]
In OPA,
OAP + OPA + POA =
[using Angle sum property of a triangle]
+ POA =
+ POA =
POA =
8. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
8. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Explanation:
Presented: PQ is a circle diameter with centre O.
The lines AB and CD are the tangents at P and Q correspondingly.
Required To Prove: AB CD
Proof: So OP is the radius along the point of contact and AB is tangent to the circle at P.
OPA = ……….(i)
[Any point on a circle has a tangent to the radius passing through the point of contact]
The CD is a circle’s Tangent at Q and OQ is the radius through the point of contact.
OQD = ……….(ii)
[Any point on a circle has a tangent to the radius passing through the point of contact]
By eq. (i) and (ii), OPA = OQD
Although these form a pair of equal alternate angles also,
AB CD
9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Explanation:
As the radius effectively travels through the centre of the circle and the tangent at any point of a circle is perpendicular to the radius through the point of contact, the perpendicular at the point of contact to the tangent to a circle passes through the centre.
10. The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
10. The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Explanation:
Any point on a circle has a tangent to the radius passing through the point of contact
OPA =
[By Pythagoras theorem]
= 9
OP = 3cm.
11. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
11. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Explanation:
Consider the O to be a similar centre of the two concentric circles.
Consider The larger circle's chord AB should touch the smaller circle at P.
Combine OP and OA.
Then, OPA =
[Any point on a circle has a tangent to the radius passing through the point of contact]
OA2 = OP2 + AP2
[By Pythagoras theorem]
= 16
AP = 4 cm
As the chord is split in half by the perpendicular from a circle's centre, it follows that
AP = BP = 4 cm
AB = AP + BP
= AP + AP = 2AP
= = 8 cm
12. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:
AB + CD = AD + BC
12. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:
AB + CD = AD + BC
Explanation:
We are conscious of the equality of the tangents to a circle from an outside point.
AP = AS ……….(i)
BP = BQ ……….(ii)
CR = CQ ……….(iii)
DR = DS……….(iv)
On Combining eq. (i), (ii), (iii) and (iv), we get
(AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS)
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
13. In the figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’Y’ at B. Prove that AOB =
13. In the figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’Y’ at B. Prove that AOB =
Explanation:
Given: In the above figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another
tangent AB with the point of contact C intersecting XY at A and X’Y’ at B.
Required To Prove: AOB =
Construction: Join OC
Proof: OPA = ……….(i)
OCA = ……….(ii)
[Any point on a circle has a tangent to the radius passing through the point of contact]
In right-angled triangles OPA and OCA,
OA = OA [Common]
AP = AC [Tangents drawn from an external
point to a circle are equal]
OPA OCA
[using RHS congruence criterion]
OAP = OAC [By C.P.C.T.]
OAC = PAB ……….(iii)
In the same way, OBQ = OBC
OBC = QBA ……….(iv)
XY X’Y’ and a transversal AB intersect them.
PAB + QBA =
[The cumulative interior angle sums on the same side of the transversal is ]
PAB + QBA
= ……….(v)
OAC + OBC =
[From eq. (iii) & (iv)]
In AOB,
OAC + OBC + AOB =
[Angle sum property of a triangle]
+ AOB = [From eq. (v)]
AOB =
Therefore proved.
14. 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.
14. 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.
Explanation:
OPA = ……….(i)
OCA = ……….(ii)
[Any point on a circle has a tangent to the radius passing through the point of contact]
OAPB is the quadrilateral
APB + AOB + OAP + OBP =
[Angle sum property of a quadrilateral]
APB + AOB + + =
[From eq. (i) & (ii)]
APB + AOB =
APB and AOB are supplementary.
15. Prove that the parallelogram circumscribing a circle is a rhombus.
15. Prove that the parallelogram circumscribing a circle is a rhombus.
Explanation:
Presented: ABCD is a parallelogram circumscribing a circle.
Required To Prove: ABCD is a rhombus.
Proof: Since the tangents from an external point to a circle are equal.
AP = AS ……….(i)
BP = BQ ……….(ii)
CR = CQ ……….(iii)
DR = DS……….(iv)
On combining eq. (i), (ii), (iii) and (iv), we achieve
(AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS)
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = AD + AD
[As Opposite sides of gm are similar]
2AB = 2AD
AB = AD
As AB = CD and AD = BC
[As Opposite sides of gm are similar]
AB = BC = CD = AD
Parallelogram ABCD is a rhombus.
16. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
16. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
Explanation:
Combine OE and OF. Also Combine OA, OB and OC.
As BD = 8 cm
BE = 8 cm
[Tangents from an exterior point to a circle are equal]
Since CD = 6 cm
CF = 6 cm
[Tangents from an exterior point to a circle are equal]
Let AE = AF =
Since OD = OE = OF = 4 cm
[Radii of a circle are similar]
Semi-perimeter of ABC = = cm
Area of ABC =
=
= cm2
Now, Area of ABC = Area of OBC + Area of OCA + Area of OAB
=
=
=
=
Applying Squaring on both sides,
AB = = 7 + 8 = 15 cm
And AC = = 7 + 6 = 13 cm
17. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
17. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Explanation:
Presented: ABCD is a quadrilateral circumscribing a circle whose centre is O.
Required To prove: (i) AOB + COD = (ii) BOC + AOD =
Construction: Join OP, OQ, OR and OS.
Proof: Because the tangents to a circle from an outside point are equal.
AP = AS,
BP = BQ ……….(i)
CQ = CR
DR = DS
In OBP and OBQ,
OP = OQ [The circle having equal Radii]
OB = OB [Common]
BP = BQ [From eq. (i)]
OPB OBQ [According to SSS congruence criterion]
[By C.P.C.T.]
In the same way,
As the sum of all the angles around a point is equal to
AOB + COD =
In the same way, we can prove that
BOC + AOD =
Also Read: Circles Class 10 Extra Questions
Chapter 10 CIRCLES