Circle+Theorems

Circle Theorems

=Proof of Theorems:=

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 =Circle Theorems=

Circles
A circle is a set of points which are all a certain distance from a fixed point known as the centre. A line joining the centre of a circle to any of the points on the circle is known as a //radius//. The //circumference// of a circle is the length of the circle. The circumference of a circle = 2 × π × the radius.

The red line in the second diagram is called a chord. It divides the circle into a major segment and a minor segment.

Angles Subtended on the Same Arc
Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.

Angle in a Semi-Circle
Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So **c is a right angle**.

[[image:http://www.mathsrevision.net/gcse/higher.GIF caption="This proof is higher tier"]] Proof
We can split the triangle in two by drawing a line from the centre of the circle to the point on the circumference our triangle touches. We know that each of the lines which is a radius of the circle (the green lines) are the same length. Therefore each of the two triangles is isosceles and has a pair of equal angles. But all of these angles together must add up to 180°, since they are the angles of the original big triangle. Therefore x + y + x + y = 180, in other words 2(x + y) = 180. and so x + y = 90. But x + y is the size of the angle we wanted to find.

Tangents
A tangent to a circle is a straight line which touches the circle at only one point (so it does not cross the circle- it just touches it). A tangent to a circle forms a right angle with the circle's radius, at the point of contact of the tangent. Also, if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.

Angle at the Centre
The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. **a = 2b**.

[[image:http://www.mathsrevision.net/gcse/higher.GIF caption="This proof is higher tier"]] Proof
You might have to be able to prove this fact: OA = OX since both of these are equal to the radius of the circle. The triangle AOX is therefore isosceles and so ∠OXA = a Similarly, ∠OXB = b

Since the angles in a triangle add up to 180, we know that ∠XOA = 180 - 2a Similarly, ∠BOX = 180 - 2b Since the angles around a point add up to 360, we have that ∠AOB = 360 - ∠XOA - ∠BOX = 360 - (180 - 2a) - (180 - 2b) =2a + 2b= 2(a + b) = 2 ∠AXB

[[image:http://www.mathsrevision.net/gcse/higher.GIF caption="This section is higher tier"]] Alternate Segment Theorem
This diagram shows the **alternate segment theorem**. In short, the red angles are equal to each other and the green angles are equal to each other.

Proof
You may have to be able to prove the alternate segment theorem: We use facts about [|related angles]: A tangent makes an angle of 90 degrees with the radius of a circle, so we know that ∠OAC + x = 90. The angle in a semi-circle is 90, so ∠BCA = 90. The [|angles in a triangle] add up to 180, so ∠BCA + ∠OAC + y = 180 Therefore 90 + ∠OAC + y = 180 and so ∠OAC + y = 90 But OAC + x = 90, so ∠OAC + x = ∠OAC + y Hence x = y



We've drawn a couple of radii (i.e. lines from the centre of the circle, we've made them red for clarity) to the ends of the chord. This makes an isosceles triangle with two equal angles which we've marked "x". We split the isosceles triangle into two identical little right angled triangles. Each of them has an angle y at the centre of the circle. The angle between the radius and the tangent is 90º (see page 114). Now we just play with the angles a bit: > then y + x =90.
 * As the angle between a tangent and a radius is 90º then angles a + x = 90.
 * As the two small angles in a right angled triangle add up to 90º
 * If a + x =90 and y + x = 90, therefore a = y.
 * We need to use the theorem "The angle at the centre is twice the angle at the circumference" (see page 111). The angle at the centre is 2y, so the angle b at the circumference = y.
 * If b = y and a = y, therefore a = b. We've done it!


 * [[image:http://www.murderousmaths.co.uk/mystw1.gif width="41" height="120" caption="Go THIS way" link="http://www.murderousmaths.co.uk/books/bkmmDB.htm"]] || [[image:http://www.murderousmaths.co.uk/mystw2.gif width="38" height="120" caption="Go THAT way" link="http://www.murderousmaths.co.uk/books/bkmm6xPT.htm"]] ||

Cyclic Quadrilaterals
A **cyclic quadrilateral** is a four-sided figure in a circle, with each vertex (corner) of the quadrilateral touching the circumference of the circle. The opposite angles of such a quadrilateral add up to 180 degrees.

[[image:http://www.mathsrevision.net/gcse/higher.GIF caption="This proof is higher tier"]] Area of Sector and Arc Length
If the radius of the circle is r, Area of sector = πr2 × A/360 Arc length = 2πr × A/360

In other words, area of sector = area of circle × A/360 arc length = circumference of circle × A/360