Looking for complete, step-by-step textbook solutions for SEBA Class 10 Mathematics Chapter 10 (Circles – Exercise 10.1) for the 2026–27 academic session? Access verified solutions to all textual questions and fill-in-the-blanks based on circle fundamentals and Theorem 10.1 (the tangent at any point of a circle is perpendicular to the radius through the point of contact). Master foundational concepts including how many tangents a circle can have, secant vs tangent intersections, point of contact definition, solving right-triangle Pythagoras problems (finding the length of tangent PQ with radius OP = 5 cm and OQ = 12 cm), and model answers structured strictly according to the Assam Board (SEBA) HSLC curriculum.
1. How many tangents can a circle have?
Answer: Infinitely many
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
(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
Soln:
Given:
Radius of circle, OP = 5 cm Distance from centre, OQ = 12 cm
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact: OP ⊥ PQ, which means ∠OPQ = 90°
In right-angled triangle OPQ, by Pythagoras theorem:
OQ² = OP² + PQ²
12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 144 – 25
PQ² = 119
PQ = √119 cm
Correct Option: (D) √119 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.
Soln:
Steps of Construction:
- Draw a circle with centre O.
- Draw a given straight line ‘l’ outside the circle.
- Draw a line ‘m’ parallel to line ‘l’ such that it touches the circle at exactly one point P. Line ‘m’ is the tangent to the circle.
- Draw a line ‘n’ parallel to line ‘l’ such that it intersects the circle at two distinct points A and B. Line ‘n’ is the secant to the circle.
SEBA Class 10 Maths Exercise 10.1 Solutions
5. In figure 10.6, the length of the tangent AP, from an external point A is 24 cm. If the distance of the point A from the centre O of the circle is 25cm, then the diameter of the circle is
(A) 15 cm
(B) 14 cm
(C) 7 cm
(D) 12 cm
Soln: Given:
Length of tangent, AP = 24 cm Distance of point A from centre O, OA = 25 cm
Since the radius is perpendicular to the tangent at the point of contact:
OP ⊥ AP, so ∠OPA = 90°
In right triangle OPA, by Pythagoras theorem:
OA² = OP² + AP²
25² = OP² + 24²
625 = OP² + 576
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm
Radius (r) = 7 cm
Diameter = 2 × r = 2 × 7 cm = 14 cm
Correct Option: (B) 14 cm
6. In figure 10.7, two circles with centres O and O’ intersect each other at A and B. AB intersects OO’ at M, then ∠OMA is
(A) 50°
(B) 60°
(C) 180°
(D) 90°
Soln: Given:
Two circles with centres O and O’ intersect at points A and B. The common chord is AB and the line segment joining the centres is OO’. AB intersects OO’ at M.
Property:
The line segment joining the centres of two intersecting circles is the perpendicular bisector of their common chord.
Therefore:
OO’ ⊥ AB
⇒ ∠OMA = 90°
Correct Option: (D) 90°
Class 10 Maths Chapter 10.1 Circles SEBA 2026-27
7. Assertion (A) : The length of the diameter of the circle with centre O is 6 cm. PT is a tangent from an external point P to the circle at the pt T. Distance of point P and the centre O is 5cm. Then the length of the tangent PT is 4 cm. (fig. 10.8)
Reason (R) : The radius through the point of contact of tangent to a circle is perpendicular to the tangent.
(A) Assertion (A) and Reason (R) both are true, Reason (R) is the correct explanation for Assertion (A)
(B) Assertion (A) and Reason (R) both are true, Reason (R) is not the correct explanation for Assertion (A)
(C) Assertion (A) is true but Reason (R) is false
(D) Assertion (A) is false but Reason (R) is true
Soln:
Verification of Assertion (A):
Diameter = 6 cm ⇒ Radius (r) = OT = 6 / 2 = 3 cm
Distance from centre, OP = 5 cm
Since radius is perpendicular to tangent at point of contact:
OT ⊥ PT ⇒ ∠OTP = 90°
In right-angled triangle OTP:
OP² = OT² + PT²
5² = 3² + PT²
25 = 9 + PT²
PT² = 16
PT = 4 cm
Therefore, Assertion (A) is true.
Verification of Reason (R):
By standard theorem, the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
Therefore, Reason (R) is true.
Since Reason (R) correctly provides the geometric basis used to apply Pythagoras theorem in Assertion (A): Reason (R) is the correct explanation for Assertion (A).
Correct Option: (A) Assertion (A) and Reason (R) both are true,
Reason (R) is the correct explanation for Assertion (A)
8. The line segment joining the points of contact of two parallel tangents to a circle is a/an
(i) radius of the circle
(ii) diameter of the circle
(iii) secant of the circle
(iv) arc of the circle
Which of the above are correct?
(A) Only (ii)
(B) (i) and (ii)
(C) Only (iii)
(D) (i), (ii) and (iv)
Soln:
Explanation:
Two parallel tangents to a circle can only be drawn at the opposite ends of a diameter. Therefore, the line segment joining their points of contact passes through the centre of the circle and equals twice the radius, which is the diameter of the circle.
Correct Option: (A) Only (ii)
SEBA Class 10 Maths Ex 10.1 Question Answer
9. If a chord of a circle is equal to its radius then the angle subtended by the chord at the centre of the circle is
(A) 30°
(B) 60°
(C) 45°
(D) 90°
Soln:
Let the chord be AB and the centre of the circle be O. Radius = OA = OB
Given: Length of chord AB = Radius = OA = OB
Therefore, in triangle OAB:
OA = OB = AB
Triangle OAB is an equilateral triangle.
Every angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Correct Option: (B) 60°
Class 10 Circles Tangent Theorem 10.1 Step by Step Solutions
10. If PT is tangent drawn from P to a circle, touching it at T and if O is the centre of the circle. Find the sum of the measures of ∠OPT and ∠POT.
Soln:
Given:
PT is tangent to the circle at T, and O is the centre.
Since the radius through the point of contact is perpendicular to the tangent:
OT ⊥ PT
⇒ ∠OTP = 90°
In triangle OPT, the sum of all interior angles is 180°:
∠OPT + ∠POT + ∠OTP = 180°
∠OPT + ∠POT + 90° = 180°
∠OPT + ∠POT = 180° – 90°
∠OPT + ∠POT = 90°
Answer:
90°
SEBA HSLC Class 10 Maths Chapter 10 Exercise 10.1 Solved PDF
⭕ HSLC Class 10 Maths Exercise 10.1 Notice: This page features the complete, step-by-step SEBA Class 10 Mathematics Chapter 10 (Circles) Exercise 10.1 Official Textual Solutions updated for the 2026–27 academic session. Formatted strictly according to the HSLC board evaluation standard, every conceptual definition, fill-in-the-blank prompt, and numerical calculation applying Theorem 10.1 OP is perpendicular to PQ and the Pythagorean theorem OQ^2 = OP^2 + PQ^2 is broken down clearly to ensure you secure full marks in your 1-mark and 2-mark board exam questions.
💡 Master SEBA Class 10 Circles:
- Having doubts about the difference between a secant and a tangent, how two parallel tangents touch opposite ends of a diameter, or solving for √119 cm (or sqrt(119) cm) in Question 3? Leave your question in the comments section below!
- Bookmark this page for quick theorem revision before your unit tests, pre-board exams, and the upcoming HSLC examinations.
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