Chapter 10: Circles
Exercise & In-Text Questions & Answers

Answer & Reference:
A circle can have infinitely many tangents. This is because a tangent can be drawn at every point on the circumference of the circle, and there are infinitely many points on a circle. (Section 10.1)

Answer & Reference:
(i) one
(ii) secant
(iii) two
(iv) point of contact
(Section 10.1, Page 209)

Answer & Reference:
(D) √119 cm
Explanation: In ΔOPQ, OP is the radius (5 cm), OQ = 12 cm, and ∠OPQ = 90° (tangent is perpendicular to radius).
By Pythagoras theorem: PQ² = OQ² - OP² = 12² - 5² = 144 - 25 = 119
Therefore, PQ = √119 cm. (Theorem 10.1, Page 210)

Answer & Reference:
The diagram below shows a circle with a given line L. Line M is parallel to L and is a tangent to the circle (touching at one point). Line N is parallel to L and is a secant (intersecting at two points). (Section 10.1, Fig. 10.1)

Answer & Reference:
(A) 7 cm
Explanation: Let O be the centre, P be the point of contact, and Q be the external point.
PQ = 24 cm (tangent), OQ = 25 cm
In right triangle OPQ, by Pythagoras theorem:
OP² = OQ² - PQ² = 25² - 24² = 625 - 576 = 49
Therefore, OP = √49 = 7 cm. (Theorem 10.2, Page 212)

Answer & Reference:
(B) 70°
Explanation: In quadrilateral OPTQ:
∠OPT = ∠OQT = 90° (tangents perpendicular to radii)
∠POQ = 110° (given)
Sum of angles in quadrilateral = 360°
∠PTQ = 360° - (90° + 90° + 110°) = 70°
(Theorem 10.2, Page 212)

Answer & Reference:
(A) 50°
Explanation: In quadrilateral OAPB:
∠APB = 80° (given)
∠OAP = ∠OBP = 90° (tangents perpendicular to radii)
∠AOB = 360° - (90° + 90° + 80°) = 100°
Since triangles OAP and OBP are congruent, ∠POA = ∠POB = 100°/2 = 50°
(Theorem 10.2, Page 212)

Answer & Reference:
Proof: Let AB be a diameter of a circle with centre O. Let PQ and RS be tangents at points A and B respectively.
Since tangent at any point is perpendicular to the radius through the point of contact (Theorem 10.1):
∠OAP = 90° and ∠OBR = 90°
Now, AB is a straight line (diameter), so ∠OAP and ∠OBR are alternate interior angles for lines PQ and RS with transversal AB.
Since ∠OAP = ∠OBR = 90°, PQ || RS (by converse of alternate interior angles theorem).
(Theorem 10.1, Page 210)

Answer & Reference:
Proof: Let XY be a tangent to a circle with centre O at point P. Let PQ be perpendicular to XY at P.
We know that the tangent at any point is perpendicular to the radius through the point of contact (Theorem 10.1).
Therefore, OP ⊥ XY.
But PQ is also perpendicular to XY at P.
Since only one perpendicular can be drawn to a line at a given point, OP and PQ must be the same line.
Hence, PQ passes through the centre O.
(Theorem 10.1, Page 210)

Answer & Reference:
Solution: Let O be the centre, P be the point of contact, and A be the external point.
AP = 4 cm (tangent), OA = 5 cm
In right triangle OPA, by Pythagoras theorem:
OP² = OA² - AP² = 5² - 4² = 25 - 16 = 9
Therefore, OP = √9 = 3 cm.
Hence, the radius of the circle is 3 cm.
(Theorem 10.2, Page 212)

Answer & Reference:
Solution: Let O be the common centre. Let AB be the chord of the larger circle touching the smaller circle at P.
Since AB touches the smaller circle at P, OP ⊥ AB (Theorem 10.1).
In right triangle OPA:
OA = 5 cm (radius of larger circle), OP = 3 cm (radius of smaller circle)
AP² = OA² - OP² = 25 - 9 = 16
AP = 4 cm
Since OP ⊥ AB, it bisects AB (perpendicular from centre bisects chord).
Therefore, AB = 2 × AP = 2 × 4 = 8 cm.
(Example 1, Page 212)

Answer & Reference:
Proof: Let the circle touch sides AB, BC, CD, DA at points P, Q, R, S respectively.
We know that the lengths of tangents from an external point to a circle are equal (Theorem 10.2).
Therefore:
AP = AS (from point A)
BP = BQ (from point B)
CQ = CR (from point C)
DR = DS (from point D)
Adding these equations:
AP + BP + CQ + DR = AS + BQ + CR + DS
(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CR)
AB + CD = AD + BC
(Theorem 10.2, Page 212)

Key Concepts & Reference:

1. Meaning of a tangent to a circle: A tangent to a circle is a line that intersects the circle at only one point. (Section 10.1, Page 207)
2. Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. (Page 210)
3. Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. (Page 212)
4. Number of tangents from a point:
   • No tangent can be drawn from a point inside the circle.
   • Exactly one tangent can be drawn from a point on the circle.
   • Exactly two tangents can be drawn from a point outside the circle.
   (Section 10.3, Page 211)

Position of Point	Number of Tangents	Explanation
Inside the circle	0	All lines through the point intersect the circle at two points
On the circle	1	Only one tangent possible at the point of contact
Outside the circle	2	Two tangents can be drawn from the external point

(Section 10.3, Page 211)