CIRCLES - Q&A

1. How many tangents can a circle have?

Answer: A circle can have infinitely many tangents. Since there are infinitely many points on a circle, and at each point, a unique tangent can be drawn, the total number of tangents is infinite.

2. Fill in the blanks :

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

Answer: one

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

Answer: secant

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

Answer: two (A secant can have two parallel tangents, one at each endpoint of the diameter perpendicular to it).

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

Answer: 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) √119 cm.

Answer:
We know that the radius is perpendicular to the tangent at the point of contact.
So, OP ⊥ PQ, which means ΔOPQ is a right-angled triangle right angled at P.
Given: Radius OP = 5 cm, OQ = 12 cm.
By Pythagoras theorem in ΔOPQ:
OQ² = OP² + PQ²
12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 144 - 25
PQ² = 119
PQ = √119 cm.
Correct Option: (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.

Answer:
To perform this construction:
1. Draw a circle with any radius and center O.
2. Draw a straight line AB anywhere outside or inside the circle (this is the given line).
3. To draw a tangent parallel to AB: Draw a radius perpendicular to line AB. At the point where this radius meets the circle, draw a line perpendicular to the radius. This line touches the circle at one point and is parallel to AB.
4. To draw a secant parallel to AB: Draw any line that cuts the circle at two distinct points and is parallel to line AB.
(Visual representation would show a circle with three parallel lines: one external/internal given line, one line touching the edge, and one line cutting through the circle).

In Q.1 to 3, choose the correct option and give justification.

1. 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

Answer:
Let O be the centre of the circle and T be the point of contact.
QT is the tangent length = 24 cm.
OQ is the distance from centre = 25 cm.
Radius OT is perpendicular to tangent QT (Theorem 10.1).
In right ΔOTQ, by Pythagoras theorem:
OQ² = OT² + QT²
25² = OT² + 24²
625 = OT² + 576
OT² = 625 - 576
OT² = 49
OT = 7 cm.
Correct Option: (A)

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Answer:
OP and OQ are radii drawn to the tangents TP and TQ respectively.
Therefore, OP ⊥ TP and OQ ⊥ TQ.
So, ∠OPT = 90° and ∠OQT = 90°.
In quadrilateral POQT, the sum of angles is 360°.
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
∠PTQ + 90° + 110° + 90° = 360°
∠PTQ + 290° = 360°
∠PTQ = 360° - 290° = 70°.
Correct Option: (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°, then ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°

Answer:
We are given that the angle between tangents PA and PB is 80°, so ∠APB = 80°.
In ΔOAP and ΔOBP:
PA = PB (Tangents from an external point are equal)
OA = OB (Radii of the same circle)
OP = OP (Common side)
Therefore, ΔOAP ≅ ΔOBP (SSS congruence criterion).
This implies ∠APO = ∠BPO = 1/2 ∠APB = 1/2 × 80° = 40°.
Since OA is perpendicular to PA (Radius ⊥ Tangent), ∠OAP = 90°.
In right ΔOAP:
∠POA + ∠OAP + ∠APO = 180°
∠POA + 90° + 40° = 180°
∠POA + 130° = 180°
∠POA = 50°.
Correct Option: (A)

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

Solution:
Let AB be a diameter of a circle with centre O.
Let two tangents PQ and RS be drawn at points A and B respectively.
Since OA is the radius to tangent PQ, OA ⊥ PQ. Therefore, ∠OAP = 90°.
Since OB is the radius to tangent RS, OB ⊥ RS. Therefore, ∠OBS = 90°.
Since AB is a straight line (diameter), we can consider it as a transversal intersecting lines PQ and RS.
∠OAP and ∠OBS are consecutive interior angles (or we can look at alternate interior angles like ∠OAQ and ∠OBS if we orient the lines differently).
Let's use alternate interior angles: ∠PAO = 90° and ∠SBO = 90° (angles on opposite sides of the transversal AB).
Wait, strictly speaking, PQ ⊥ AB and RS ⊥ AB.
Since both lines PQ and RS are perpendicular to the same line AB, they must be parallel to each other.
Therefore, PQ || RS.

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

Solution:
Let AB be a tangent to a circle at point P. Let O be the centre.
We know that the radius is perpendicular to the tangent at the point of contact (Theorem 10.1).
So, OP ⊥ AB.
Therefore, the line segment perpendicular to the tangent AB at P is the line containing the radius OP.
Since the radius connects the point on the circle to the centre, the line containing the radius must pass through the centre O.
Alternative Method (Contradiction):
Assume the perpendicular at P does NOT pass through the centre O. Let it pass through some other point O'.
Then ∠O'PB = 90° (by assumption).
But we know radius OP ⊥ tangent AB, so ∠OPB = 90°.
This implies ∠O'PB = ∠OPB, which is only possible if the line O'P coincides with the line OP.
Therefore, the perpendicular must pass through the centre O.

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

Solution:
Let O be the centre and T be the point of contact of the tangent from A.
OA = 5 cm (Distance from centre).
AT = 4 cm (Length of tangent).
Since radius OT ⊥ tangent AT, ΔOTA is a right-angled triangle.
By Pythagoras theorem:
OA² = OT² + AT²
5² = OT² + 4²
25 = OT² + 16
OT² = 25 - 16 = 9
OT = 3 cm.
Answer: The radius of the circle is 3 cm.

7. 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.

Solution:
Let O be the common centre of the two concentric circles.
Let AB be the chord of the larger circle that touches the smaller circle at point P.
Since AB is a tangent to the smaller circle at P, radius OP ⊥ AB.
We are given:
Radius of larger circle (OA) = 5 cm.
Radius of smaller circle (OP) = 3 cm.
In right-angled ΔOPA:
OA² = OP² + AP²
5² = 3² + AP²
25 = 9 + AP²
AP² = 16
AP = 4 cm.
Since the perpendicular from the centre to a chord bisects the chord, P is the mid-point of AB.
Therefore, Length of chord AB = 2 × AP = 2 × 4 = 8 cm.
Answer: The length of the chord is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.

Solution:
Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
We know that lengths of tangents drawn from an external point to a circle are equal.
Therefore:
From A: AP = AS ...(1)
From B: BP = BQ ...(2)
From C: CR = CQ ...(3)
From D: DR = DS ...(4)

Adding equations (1), (2), (3), and (4):
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
Using the figure, AP + BP = AB, CR + DR = CD, AS + DS = AD, BQ + CQ = BC.
Substituting these values:
AB + CD = AD + BC
Hence Proved.

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 ∠ AOB = 90°.

Solution:
Join O to point C.
In ΔOPA and ΔOCA:
OP = OC (Radii)
AP = AC (Tangents from A)
OA = OA (Common)
So, ΔOPA ≅ ΔOCA (SSS congruence).
Therefore, ∠POA = ∠COA. Let these be x. So ∠POC = 2x.
Similarly, ΔOQB ≅ ΔOCB.
Therefore, ∠QOB = ∠COB. Let these be y. So ∠QOC = 2y.

Since POQ is a straight line (diameter) passing through the centre perpendicular to parallel tangents:
Angle on straight line POQ = 180°
∠POA + ∠COA + ∠COB + ∠QOB = 180°
x + x + y + y = 180°
2x + 2y = 180°
2(x + y) = 180°
x + y = 90°
From the figure, ∠AOB = ∠COA + ∠COB = x + y.
Therefore, ∠AOB = 90°.
Hence 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:
Let PA and PB be two tangents from external point P to a circle with centre O. A and B are points of contact.
We need to prove that ∠APB + ∠AOB = 180°.
In quadrilateral OAPB:
Angle between radius and tangent is 90°.
So, ∠OAP = 90° and ∠OBP = 90°.
The sum of angles in a quadrilateral is 360°.
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + ∠AOB = 360°
180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° - 180°
∠APB + ∠AOB = 180°
Hence, the angles are supplementary.

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

Solution:
Let ABCD be a parallelogram circumscribing a circle.
From the property proved in Question 8 (opposite sides sum is equal):
AB + CD = AD + BC.
Since ABCD is a parallelogram, opposite sides are equal.
AB = CD and AD = BC.
Substituting these into the equation:
AB + AB = AD + AD
2AB = 2AD
AB = AD.
Since adjacent sides of the parallelogram are equal (AB = AD), all sides are equal (AB = BC = CD = DA).
A parallelogram with all sides equal is a rhombus.
Therefore, ABCD is a rhombus.

12. 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 Fig. 10.14). Find the sides AB and AC.

Solution:
Let the circle touch AC at E and AB at F.
Given radius OD = OE = OF = 4 cm.
Tangents from an external point are equal:
CD = CE = 6 cm.
BD = BF = 8 cm.
Let AF = AE = x cm.

Now the sides of the triangle are:
a = BC = 6 + 8 = 14 cm
b = AC = 6 + x cm
c = AB = 8 + x cm

Calculate Area of ΔABC using Heron's Formula:
Semi-perimeter s = (14 + (6 + x) + (8 + x)) / 2 = (28 + 2x) / 2 = 14 + x.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[(14+x) (14+x - 14) (14+x - (6+x)) (14+x - (8+x))]
Area = √[(14+x) (x) (8) (6)]
Area = √[48x(14+x)] ...(1)

Calculate Area of ΔABC using 3 smaller triangles (ΔOBC, ΔOCA, ΔOAB):
Area = Area(ΔOBC) + Area(ΔOCA) + Area(ΔOAB)
Area = (1/2 × BC × OD) + (1/2 × AC × OE) + (1/2 × AB × OF)
Area = (1/2 × 14 × 4) + (1/2 × (6+x) × 4) + (1/2 × (8+x) × 4)
Area = 28 + 2(6+x) + 2(8+x)
Area = 28 + 12 + 2x + 16 + 2x
Area = 56 + 4x = 4(14 + x) ...(2)

Equating (1) and (2):
√[48x(14+x)] = 4(14+x)
Squaring both sides:
48x(14+x) = 16(14+x)²
Divide by 16(14+x) (Since x > 0, 14+x is not zero):
3x = 14 + x
2x = 14
x = 7 cm.

Now find sides AB and AC:
AB = 8 + x = 8 + 7 = 15 cm.
AC = 6 + x = 6 + 7 = 13 cm.
Answer: AB = 15 cm, AC = 13 cm.

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

Solution:
Let ABCD be the quadrilateral circumscribing a circle with centre O.
Let the points of contact be P, Q, R, S on sides AB, BC, CD, DA respectively.
Join the centre O to the vertices A, B, C, D and points of contact P, Q, R, S.
We have 8 small triangles formed at the centre.
Compare adjacent triangles ΔOAP and ΔOAS:
AP = AS (Tangents)
OP = OS (Radii)
OA = OA (Common)
So ΔOAP ≅ ΔOAS. Thus angles at centre are equal: ∠1 = ∠8 (labeling angles sequentially 1 to 8 around the centre starting from POA).
Similarly, ∠2 = ∠3 (for ΔOBP and ΔOBQ)
∠4 = ∠5 (for ΔOCQ and ΔOCR)
∠6 = ∠7 (for ΔODR and ΔODS)

Sum of all angles at centre is 360°:
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
Substitute equalities:
(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°
2∠1 + 2∠2 + 2∠5 + 2∠6 = 360° (grouping for opposite sides AB and CD)
2(∠1 + ∠2) + 2(∠5 + ∠6) = 360°
(∠1 + ∠2) + (∠5 + ∠6) = 180°
Here ∠1 + ∠2 = ∠AOB and ∠5 + ∠6 = ∠COD.
So, ∠AOB + ∠COD = 180°.
Similarly, ∠BOC + ∠DOA = 180°.
Therefore, opposite sides subtend supplementary angles at the centre.