Summary of Circle Theorems and Chords

## Introduction Circle geometry studies the relationships between the centre, radii, chords, diameters and angles in a circle. In this material we focus on important theorems about lines drawn from the centre to chords, their converses, and worked examples that show how to apply these theorems to calculate lengths in circles. > **Definition:** A *chord* is a straight line joining two points on a circle. A *radius* is a line from the centre to any point on the circle. A *diameter* is a chord that passes through the centre. ## Key Theorems and Ideas ### Theorem 1: Perpendicular from centre bisects the chord If a line from the centre of a circle is perpendicular to a chord, then it bisects the chord. > **Definition:** Bisect means divide into two equal parts. Explanation (idea broken down): - Let $O$ be the centre and let $OD$ be perpendicular to chord $AB$ at $D$. - Join $OA$ and $OB$ (these are radii). Triangles $OAD$ and $OBD$ are right triangles with two equal sides $OA=OB$ and share $OD$, and both have a right angle at $D$. - By congruence ($RHS$), $ riangle OAD\equiv\triangle OBD$, hence $AD=DB$. ### Converse of Theorem 1 A line drawn from the centre of the circle to the midpoint of a chord is perpendicular to the chord. Explanation (idea broken down): - If $D$ is the midpoint of chord $AB$ and $O$ is the centre, then $OD$ joins the centre to the midpoint. - Using similar congruence reasoning with radii and equal halves of the chord, $OD$ must be perpendicular to $AB$. ## Worked Examples with Steps ### Example 1: Bisected chord and right triangle (adapted) Given a circle with centre $O$, chord $AB=24\text{ cm}$, and $OD$ is perpendicular to $AB$ at $D$. Find $AD$ and $OD$ if $AO=20\text{ cm}$. Solution: 1. Since $OD$ is perpendicular to $AB$, then $AD=DB$. So $$AD=\frac{AB}{2}=\frac{24}{2}=12\text{ cm}.$$ 2. Consider right triangle $AOD$. Use Pythagoras: $$AO^2=AD^2+OD^2$$ $$20^2=12^2+OD^2$$ $$OD^2=400-144=256$$ $$OD=16\text{ cm}.$$ ### Example 2: Adding segment lengths (adapted) Given a circle with centre $O$, points $P,Q,R$ on a chord with $PQ=16\text{ cm}$ and $OQ$ perpendicular to the chord at $Q$. If $Q$ is midpoint of $PR$, find $PR$. Solution: - If $Q$ is the midpoint, $PQ=QR=16\text{ cm}$. Therefore $$PR=PQ+QR=16+16=32\text{ cm}.$$ ### Example 3: Right triangle from radius (adapted) Given $PO$ is a radius, $PQ=8\text{ cm}$ and $OS=PO$, $QS=2\text{ cm}$ with $S$ on the same line as $Q$. If $PO^2=PQ^2+OQ^2$ and $PO=OQ+QS$, then substitute $QS=2$ to form an equation for $OQ$ and $PO$. Work: From $PO=OQ+2$ and $PO^2=8^2+OQ^2$, substitute to get $$\left(OQ+2\right)^2=64+OQ^2.$$ Expand and simplify to solve for $OQ$. ## Problems to Try 1. A chord of length $30\text{ cm}$ is bisected by a line from the centre. If the radius is $13\text{ cm}$, find the distance from the centre to the chord. (Use Pythagoras.) 2. In a circle with centre $O$, $OD$ meets chord $AB$ at its midpoint $D$. If $OA=10\text{ cm}$ and $AD=6\text{ cm}$ find $OD$. 3. Show algebraically that if $PO^2=PQ^2+OQ^2$ and $PO=OQ+2$ with $PQ=8$, the positive solution for $OQ$ is $6$ cm. ## Real-world applications - Engineering: designing circular gears where chords represent contact points and distances to centre determine stresses. - Architecture: calculating load-bearing elements that subtend chords in circular arches. Fun fact: A circle has infinite lines of symmetry through its centre, and any line through the centre that is perpendicular to a chord will always split that chord into equal parts. ## Comparison table: Related concepts | Concept | Condition | Result | |---|---:|---| | Line from centre perpendicular to chord | $OD\perp AB$ | $AD=DB$ | | Line from centre to midpoint of chord | $D$ midpoint of $AB$ | $OD\perp AB$ | | Radius to point on circle | $OA$ | Constant length for all points | ## Summary - A perpendicular from the centre to a chord bisects the chord. The converse also holds: a line from the centre to the midpoint of a cho