Chapter 10 : Circles Class 10 Maths NCERT Solutions

Chapter 10: Circles — Exercise 10.1

Welcome to the foundation of circular geometry! Understanding how lines interact with circles is the “secret sauce” behind everything from the design of bicycle gears to the way satellites orbit our planet.

Q1. How many tangents can a circle have?

• The Logic: Think of a circle as a collection of infinite tiny points. At every single one of those points, you can draw exactly one line that just “kisses” the edge without crossing into the circle. Since there are infinite points, there are infinite possibilities!

• The Result: A circle can have infinitely many tangents.

Q2. 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 __________ .

• The Logic: These are the core definitions. A Tangent is a “touching line” (1 point), while a Secant is a “cutting line” (2 points). Parallel tangents only happen at the exact opposite ends of a diameter.

• The Steps:

  • (i) By definition, a tangent only touches one spot.

  • (ii) If it cuts through, it’s a secant.

  • (iii) Imagine sliding a ruler across a coin; it only stays parallel at the two “poles.”

  • (iv) This is the specific “meeting spot.”

• The Result:

  • (i) exactly one

  • (ii) secant

  • (iii) two

  • (iv) point of contact

Q3. 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) root 119 cm

• The Logic: This is where Theorem 10.1 comes into play. The radius and the tangent always meet at a 90-degree angle at the point of contact. This creates a perfect Right-Angled Triangle (OPQ), allowing us to use the Pythagorean Theorem.

• The Steps:

  • In triangle OPQ, Angle P = 90 degrees.

  • Use Pythagoras: OP^2 + PQ^2 = OQ^2

  • Substitute the values: 5^2 + PQ^2 = 12^2

  • 25 + PQ^2 = 144

  • PQ^2 = 144 – 25 = 119

  • PQ = root(119)

• Human Tip: Students often mistake 12 as a side and 13 as the hypotenuse (the 5-12-13 triplet). But here, 12 is the distance from the center, making it the hypotenuse! Always label your longest side carefully.

• The Result: (D) root 119 cm

Q4. 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.

• The Logic: This is a visualization task. You start with a base line and then “copy” that direction to create a line that touches the circle (tangent) and another that cuts through it (secant).

• The Steps:

  1. Draw a circle with centre O.

  2. Draw any line ‘l’ outside the circle.

  3. Draw a line parallel to ‘l’ that touches the circle at exactly one point (Tangent).

  4. Draw another line parallel to ‘l’ that passes through two points in the circle (Secant).

• Pro-Tip: In exams, use a set-square or a ruler and compass to ensure your lines are perfectly parallel!

Chapter 10: Circles — Exercise 10.2

Q6. 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.

• The Logic: This is a classic Pythagoras problem. The distance from the centre (5 cm) is your Hypotenuse, the tangent (4 cm) is one side, and the radius is the other.

• The Steps:

  • Let radius be ‘r’.

  • r^2 + (Tangent)^2 = (Distance from centre)^2

  • r^2 + 4^2 = 5^2

  • r^2 + 16 = 25

  • r^2 = 9

• The Result: The radius of the circle is 3 cm.

Q7. 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.

• The Logic: Concentric circles have the same center. The chord of the big circle acts as a Tangent to the small circle. This creates a right-angled triangle where the big radius is the hypotenuse and the small radius is the perpendicular.

• The Steps:

  1. In the small triangle: (Small Radius)^2 + (Half Chord)^2 = (Big Radius)^2

  2. 3^2 + x^2 = 5^2

  3. 9 + x^2 = 25 => x^2 = 16 => x = 4 cm.

  4. Since the perpendicular from the centre bisects the chord, total length = 2 * x.

• The Result: The length of the chord is 8 cm.

Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC.

• The Logic: Every vertex of the quadrilateral is an “external point” with two equal tangents touching the circle. We list these equal pairs and add them all up.

• The Steps:

  1. Let the points of contact be P, Q, R, S.

  2. From A: AP = AS

  3. From B: BP = BQ

  4. From C: CR = CQ

  5. From D: DR = DS

  6. Adding all: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

• The Result: AB + CD = AD + BC. (Proved)

Q9. In the figure, 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 Angle AOB = 90 degrees.

• The Logic: We use congruent triangles to show that the line from the centre to the point of contact splits the angles at the centre into equal parts.

• The Steps:

  1. Join OC. Triangle OPA is congruent to triangle OCA (by SSS or RHS).

  2. So, Angle POA = Angle COA = x.

  3. Similarly, triangle OQB is congruent to triangle OCB, so Angle QOB = Angle COB = y.

  4. Since POQ is a straight line, 2x + 2y = 180.

  5. x + y = 90.

• The Result: Angle AOB = 90 degrees. (Proved)

Q10. 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.

• The Logic: “Supplementary” means the two angles add up to 180 degrees. We just use the quadrilateral rule here!

• The Steps:

  1. Let the external point be P and points of contact be A and B.

  2. In quadrilateral OAPB, Angle OAP = 90 and Angle OBP = 90.

  3. Sum of angles = 360.

  4. 90 + 90 + Angle APB + Angle AOB = 360.

  5. Angle APB + Angle AOB = 180.

• The Result: The angles are supplementary. (Proved)