Chapter 10

Class 10 Mathematics - Chapter 10: Circles
Chapter 10: Circles
1. Tangent to a Circle & Theorem 10.1
Tangent: A line that intersects a circle at only one point is called a tangent. The point where it touches the circle is called the point of contact.

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. $$OP \perp XY$$ where $O$ is the centre, $P$ is the point of contact, and $XY$ is the tangent line.
O P X Y
Example 1.1: Calculating Tangent Length / Radius

A tangent $PQ$ at a point $P$ of a circle of radius $5\text{ cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12\text{ cm}$. Find the length $PQ$.

Solution:
By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact: $$OP \perp PQ \implies \angle OPQ = 90^\circ$$ Applying Pythagoras theorem in right $\triangle OPQ$: $$OQ^2 = OP^2 + PQ^2$$ $$12^2 = 5^2 + PQ^2$$ $$144 = 25 + PQ^2$$ $$PQ^2 = 144 - 25$$ $$PQ^2 = 119$$ $$PQ = \sqrt{119}\text{ cm}$$ The length of $PQ$ is $\sqrt{119}\text{ cm}$.

2. Number of Tangents & Theorem 10.2
Key Facts:
  • From a point inside a circle: 0 tangents can be drawn.
  • From a point on the circle: Exactly 1 tangent can be drawn.
  • From a point outside a circle: Exactly 2 tangents can be drawn.

Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. $$PQ = PR$$
P Q R O
Example 2.1: Proving Angle Properties of Tangents

If $TP$ and $TQ$ are two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, find $\angle PTQ$.

Solution:
In quadrilateral $OPTQ$:
Since radius is perpendicular to tangent at the point of contact (Theorem 10.1): $$\angle OPT = 90^\circ$$ $$\angle OQT = 90^\circ$$ The sum of interior angles in quadrilateral $OPTQ$ is $360^\circ$: $$\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^\circ$$ $$90^\circ + 110^\circ + 90^\circ + \angle PTQ = 360^\circ$$ $$290^\circ + \angle PTQ = 360^\circ$$ $$\angle PTQ = 360^\circ - 290^\circ$$ $$\angle PTQ = 70^\circ$$ Therefore, $\angle PTQ = \mathbf{70^\circ}$.

Example 2.2: Tangents Encircling a Quadrangle

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

Solution:
Let the circle touch sides $AB, BC, CD, AD$ at points $P, Q, R, S$ respectively.
By Theorem 10.2 (tangents from an external point are equal): $$AP = AS \quad \text{--- (1)}$$ $$BP = BQ \quad \text{--- (2)}$$ $$CR = CQ \quad \text{--- (3)}$$ $$DR = DS \quad \text{--- (4)}$$ Adding equations (1), (2), (3), and (4): $$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ $$AB + CD = AD + BC$$ Hence proved.