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Chapter 10: Circles

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This page provides comprehensive Chapter 10: Circles - Assertion Reason Worksheet - SJMaths. Assertion and Reason type questions for Class 10 Circles. Practice for CBSE Board Exams.

Assertion-Reason Worksheet

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Directions:

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  • (C) Assertion (A) is true but Reason (R) is false.
  • (D) Assertion (A) is false but Reason (R) is true.
  1. Question 1:
    Assertion (A): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
    Reason (R): The lengths of tangents drawn from an external point to a circle are equal.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: A is a standard theorem (Theorem 10.1). A is true.
    Step 2: R is also a standard theorem (Theorem 10.2). R is true.
    Step 3: R does not explain why the tangent is perpendicular to the radius. So B is correct.
  2. Question 2:
    Assertion (A): If the length of tangent from an external point P to a circle of radius 5 cm is 12 cm, then the distance of P from the centre is 13 cm.
    Reason (R): In a right-angled triangle, Hypotenuse$^2$ = Base$^2$ + Perpendicular$^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Radius $\perp$ Tangent. In $\Delta OTP$, $OP^2 = OT^2 + TP^2 = 5^2 + 12^2 = 25 + 144 = 169$.
    Step 2: $OP = \sqrt{169} = 13$ cm. A is true.
    Step 3: R is Pythagoras theorem, which is used to calculate the distance. R explains A.
  3. Question 3:
    Assertion (A): A parallelogram circumscribing a circle is a rhombus.
    Reason (R): The lengths of tangents drawn from an external point to a circle are equal.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Using the property of tangents (R), we can prove $AB + CD = AD + BC$.
    Step 2: For a parallelogram, $AB=CD$ and $AD=BC$. Thus $2AB = 2AD \Rightarrow AB=AD$.
    Step 3: A parallelogram with equal adjacent sides is a rhombus. R explains A.
  4. Question 4:
    Assertion (A): Two concentric circles cannot have a common tangent.
    Reason (R): A circle can have at most two parallel tangents.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: Concentric circles have the same centre but different radii. A tangent to the inner circle is a secant to the outer, and a tangent to the outer circle does not touch the inner. A is true.
    Step 2: R is true but does not explain A.
  5. Question 5:
    Assertion (A): If a chord AB subtends an angle of $60^\circ$ at the centre of a circle, then the angle between the tangents at A and B is $120^\circ$.
    Reason (R): The angle between 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.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Angle between tangents + Angle at centre = $180^\circ$.
    Step 2: Angle between tangents = $180^\circ - 60^\circ = 120^\circ$. A is true.
    Step 3: R states the supplementary property used. R explains A.
  6. Question 6:
    Assertion (A): The centre of the circle inscribed in a triangle is the point of intersection of the angle bisectors.
    Reason (R): The centre of the circle circumscribing a triangle is the point of intersection of the perpendicular bisectors of the sides.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: The incentre is equidistant from sides, lying on angle bisectors. A is true.
    Step 2: The circumcentre is equidistant from vertices, lying on perpendicular bisectors. R is true.
    Step 3: R is a correct statement but about circumcircle, not incircle. So B is correct.
  7. Question 7:
    Assertion (A): In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
    Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: The chord of the larger circle is a tangent to the smaller circle.
    Step 2: Radius is perpendicular to the tangent (R).
    Step 3: Perpendicular from centre to a chord bisects the chord. Thus, A is true and R explains the perpendicularity needed for bisection.
  8. Question 8:
    Assertion (A): From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = 2r (r is radius), then $\Delta APB$ is equilateral.
    Reason (R): $\sin 30^\circ = \frac{1}{2}$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: In right $\Delta OAP$, $\sin(\angle APO) = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2}$.
    Step 2: $\angle APO = 30^\circ$. Since tangents are equally inclined, $\angle APB = 60^\circ$.
    Step 3: Since $PA=PB$, $\Delta APB$ is isosceles with one angle $60^\circ$, making it equilateral. R explains the angle calculation.
  9. Question 9:
    Assertion (A): If a line intersects a circle at two distinct points, it is called a secant.
    Reason (R): A tangent to a circle intersects it in exactly one point.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: Definition of secant is correct. A is true.
    Step 2: Definition of tangent is correct. R is true.
    Step 3: R defines tangent, which doesn't explain the definition of secant. So B is correct.
  10. Question 10:
    Assertion (A): The perpendicular at the point of contact to the tangent to a circle passes through the centre.
    Reason (R): The radius is perpendicular to the tangent at the point of contact.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Since radius is perpendicular to tangent (R), the perpendicular drawn at the point of contact must be the radius (or line containing it).
    Step 2: The radius passes through the centre. Thus, A is true and R explains it.
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