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Circle Theorems

- Calculus
- Absolute Maxima and Minima
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- Determining Volumes by Slicing
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- Evaluating a Definite Integral
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- Exponential Functions
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- Composition
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- Coordinate Systems
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- Equation of Circles
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- Fundamentals of Geometry
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- HL ASA and AAS
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- Segment Length
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- Mechanics Maths
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- Law of Cosines in Algebra
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- Limits of Accuracy
- Linear Expressions
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- Rewriting Formulas and Equations
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- Rounding
- SAS Theorem
- SSS Theorem
- Scalar Triple Product
- Scale Drawings and Maps
- Scale Factors
- Scientific Notation
- Second Order Recurrence Relation
- Sector of a Circle
- Segment of a Circle
- Sequences
- Sequences and Series
- Series Maths
- Sets Math
- Similar Triangles
- Similar and Congruent Shapes
- Simple Interest
- Simplifying Fractions
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- Simultaneous Equations
- Sine and Cosine Rules
- Small Angle Approximation
- Solving Linear Equations
- Solving Linear Systems
- Solving Quadratic Equations
- Solving Radical Inequalities
- Solving Rational Equations
- Solving Simultaneous Equations Using Matrices
- Solving Systems of Inequalities
- Solving Trigonometric Equations
- Solving and Graphing Quadratic Equations
- Solving and Graphing Quadratic Inequalities
- Special Products
- Standard Form
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- Straight Line Graphs
- Substraction and addition of fractions
- Sum and Difference of Angles Formulas
- Sum of Natural Numbers
- Surds
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- Tables and Graphs
- Tangent of a Circle
- The Quadratic Formula and the Discriminant
- Transformations
- Transformations of Graphs
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- Triangle Rules
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- Trigonometric Identities
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- Trigonometry
- Turning Points
- Types of Functions
- Types of Numbers
- Types of Triangles
- Unit Circle
- Units
- Variables in Algebra
- Vectors
- Verifying Trigonometric Identities
- Writing Equations
- Writing Linear Equations
- Statistics
- Bias in Experiments
- Binomial Distribution
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- Bivariate Data
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- Chi Square Test for Homogeneity
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- Chi-Square Distribution
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- Comparing Two Means Hypothesis Testing
- Conditional Probability
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- Conducting a Survey
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- Confidence Interval for Population Mean
- Confidence Interval for Population Proportion
- Confidence Interval for Slope of Regression Line
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- Confidence Intervals
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- Data Analysis
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- Inference for Distributions of Categorical Data
- Inferences in Statistics
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- Survey Bias
- Transforming Random Variables
- Tree Diagram
- Two Categorical Variables
- Two Quantitative Variables
- Type I Error
- Type II Error
- Types of Data in Statistics
- Venn Diagrams

Let's look at each of the circle theorems, and then their proofs. We will then look at how to apply these theorems.

This circle theorem is illustrated below. It states that for any triangle inscribed inside the circle with all points touching the circumference and the hypotenuse as a diameter, then the angle opposite the hypotenuse will be right-angled.

Let us draw a line down from the angle opposite the hypotenuse to the centre. This will mean we have divided the triangle into two further triangles, each isosceles, with two sides of length r (we use r to denote radius length). This means each triangle has two angles which are the same. We can draw this below:

Looking at the largest triangle, we know that 2x + 2y = 180° as the angles must sum to 180 °. As 2x + 2y = 180°, it follows – by dividing by two – that x + y = 90°. The angle at the circumference is given by x + y, and thus, the angle is right-angled. QED

Here, the angle subtended by an arc at the centre is twice the angle subtended at the circumference. This is shown below. What is important to note is that it doesn't matter where the point is on the arc, as long as it is between the two unmarked angles. If this happens then the theorem will still hold.

Let us construct the same shape, but now also construct a line from the 'x' point to the centre. This gives us two isosceles triangles with two sides of length r, and two sides of the same length. We will also have two angles in each isosceles that are the same. We will label each angle, as is shown below.

To prove the theorem, we need to show that 2 (a + b) = c.

Using the facts that there are 180 ° in a triangle, and 360 ° around a point, we can form three equations: 2a + z = 180, 2b + t = 180 and z + t + c = 360 ° .

We can rearrange the first equation to z = 180-2a, and rearrange the second equation to t = 180 ° -2b.

Now we can substitute these equations into the third equation, to get 180 ° -2a + 180 ° -2b + c = 360 ° .

This simplifies to c-2a-2b = 0, which can then be further simplified to 2 (a + b) = c. QED

If we have two triangles inside a circle with all three corners touching the circle, and the triangles share a side (also known as a common chord) then the third angle is the same in both triangles, as long as these third angles are in the same segment. This is shown below.

Let us first start by drawing another triangle sharing the common chord, however, this time we will connect the line to the centre. This is the same shape as seen in theorem 2, meaning we can invoke this theorem and call the angle 2x. This is shown below.

As we have used theorem 2, it doesn't matter where we put the angle x on the arc, meaning the theorem is now proved. QED

By cyclic quadrilateral, we mean a four-sided shape, all corners of which touch a circle. When this occurs, the opposite corners in the quadrilateral will sum to 180 °.

In this case, we would have a + c = 180 °, as well as b + d = 180 °.

Let us draw a line from each corner to the centre. As this goes from the circle to the centre, this means that this is a radius which means we have created four isosceles triangles which have pairs of matching angles. This is shown below.

To show that opposite angles sum to 180 °, then we must show x + y + z + t = 180 °.

The angles in a quadrilateral sum to 360 °.

This means that z + y + z + t + t + x + x + y = 360 °.

This can be simplified to 2 (x + y + z + t) = 360 °, so x + y + z + t = 180 °. QED

Suppose we drew a tangent to a circle. At the point at which the tangent touches the circle, there is a corner of a triangle. The other two corners of the triangle also lie on the circle. In this case, the angle between the tangent and the triangle is equal to the adjacent angle in the triangle. This is shown below.

To prove this you only need to show this on one side, as it does not matter what triangle we choose. Construct a triangle as above, and then join each corner to the centre. Again, we have created three isosceles triangles – all with a pair of corresponding angles. We will call the angle between the tangent and triangle a. This is all shown below.

Our aim is to show that a = z + y.

As the radius is perpendicular to the tangent at the point the tangent touches the circle (by definition), we know a + x = 90 °.

As the angles in a triangle sum to 180 °, we know 2x + 2y + 2z = 180 °, so x + y + z = 90 °.

We can rewrite our first equation as x = 90° - a, and then substitute this into the second equation, to get 90 ° -a + y + z = 90 °, which we can rearrange to a = z + y

This was our aim. QED

Find x.

By our theorem 2, we know that the angle BOC will be 2 * 70 ° = 140 °.

As the line OD bisects this angle, we know the angle DOC is 70 °.

As OC is a radius, and DC is a tangent at C, then OC is perpendicular to DC, and so angle OCD is 90 °.

This means that we can now find x. As the angles in a triangle sum to 180,

we have 90 ° + 70 ° + x = 180 °, giving x as 20 °.

x + y + z = 260. Find x, y and z

By Theorem 2, y = 2x, and by cyclic quadrilaterals, we get x + z = 180 °, which can be rearranged to z = 180 ° - x. We can then fill these into our original equation, to get x + 2x + 180 - x = 260. This simplifies to 2x = 80 °, giving x = 40 °, then y = 80 ° and z = 140 °.

Suppose we draw a circle and choose any point in the same plane as the circle, so long as the point is outside of the circle. Then, we can draw two tangent lines from the point to the circle. In addition to this, the distance from the point to the circle will be the same in both cases.

So in this case, the distance AP is the same as the distance AQ.

Let us draw lines from the tangent point to the centre, recalling that the radius at a tangent point is perpendicular to the tangent, which then gives us:

We can now use Pythagoras' theorem as we have right-angled triangles.

This gives us and

Equating the two expressions for , we arrive at:

Length is always positive so the negative solutions are ignored.

**Theorem 7: A radius which is perpendicular through a chord bisects the chord**

Suppose we have any chord in a circle, and we draw a line from the radius to the circle boundary, and this line is perpendicular to the chord. In this case, the radius will bisect the chord.

Let us draw a line from O to M and also from O to N.

For this to be a bisection, we need AN to be the same length AM.

As we have right-angled triangles, then we can use Pythagoras' theorem.

As length is positive, then AM = AN. QED

Know what the seven circle theorems are, and how to prove them

Learn how to apply these theorems to exam level problems

More about Circle Theorems

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