Class 10 Maths Chapter 10 PDF: Circles

Class 10 Maths Chapter 10, Circles, studies the interaction between a line and a circle—non-intersecting line, secant (two points) and tangent (one point of contact)—and proves the two foundational tangent theorems used throughout circle geometry.

What Class 10 Maths Chapter 10 Covers

10.1 & 10.2 Tangent to a Circle (Theorem 10.1) defines a tangent as a line intersecting the circle at only one point (the point of contact) and proves Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact ($OP \perp XY$).

10.3 Number of Tangents from a Point on a Circle (Theorem 10.2) shows there is no tangent from an interior point, exactly one tangent at a point on the circle, and exactly two tangents from an exterior point $P$, then proves Theorem 10.2 (by RHS congruence of $\triangle OPQ$ and $\triangle OPR$): the lengths of tangents drawn from an external point to a circle are equal ($PQ = PR$), and the centre lies on the bisector of $\angle QPR$.

Exercises in Chapter 10

  • Exercise 10.1 (4 Questions) covers how many tangents a circle can have (infinitely many), properties of secants and tangents, and finding tangent length using Pythagoras’s theorem.
  • Exercise 10.2 (13 Questions) applies Theorems 10.1 and 10.2 to concentric circles (chord of larger circle touching smaller circle is bisected at point of contact), quadrilaterals circumscribing a circle ($AB + CD = AD + BC$, proving a parallelogram circumscribing a circle is a rhombus), and proving opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre.

Read Class 10 Maths Chapter 10 Online

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FAQ

State Theorem 10.1 and Theorem 10.2 on tangents to a circle.
Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

If a quadrilateral $ABCD$ is drawn to circumscribe a circle, how are its sides related?
Because tangents from each exterior vertex to the circle have equal lengths ($AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$), adding them gives $AB + CD = AD + BC$ (the sums of opposite sides are equal).

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